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Multiple Choice Questions
One and More than One Correct
1. The graph between $1/\lambda$ and stopping potential ($V$) of three metals having work functions $\phi_1$, $\phi_2$ and $\phi_3$ in an experiment of photo-electric effect is plotted as shown in the figure. Which of the following statement(s) is/are correct? [JEE 2006]
[Here, $\lambda$ is the wavelength of the incident ray].
a. Ratio of work functions $\phi_1 : \phi_2 : \phi_3 = 1:2:4$
b. Ratio of work functions $\phi_1 : \phi_2 : \phi_3 = 4:2:1$
c. Slope is directly proportional to $hc/e$, where $h$ is Planck’s constant and $c$ is the speed of light.
d. The violet colour light can eject photoelectrons from metals 2 and 3.
Answer: (a, c)
Solution:
For plate 1: $\phi_1 = hc(0.001)$
Plate 2: $\phi_2 = hc(0.002)$
Plate 3: $\phi_3 = hc(0.004)$
Ratio of work functions $\phi_1 : \phi_2 : \phi_3 = 1:2:4$
For plate 2, threshold wavelength $\lambda_2 = \frac{hc}{\phi_2} = \frac{hc}{0.002 hc} = 500\text{ nm}$
For plate 3, threshold wavelength $\lambda_3 = \frac{hc}{\phi_3} = \frac{hc}{0.004 hc} = 250\text{ nm}$
Since violet colour light is $400\text{ nm}$, so $\lambda_{\text{threshold}} < \lambda_{\text{light}}$ for plate 2.
So, violet colour light will eject photo-electron form plate 2 and not form plate 3.
2. Electrons with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is: [JEE 2007 P-II]
a. $\lambda_0 = \frac{2mc\lambda^2}{h}$
b. $\lambda_0 = \frac{h}{mc}$
c. $\lambda_0 = \frac{mc\lambda^2}{2h}$
d. $\lambda_0 = \lambda$
Answer: (a)
Solution:
$\lambda = \frac{h}{mv} \implies v = \frac{h}{m\lambda}$
$\text{Kinetic Energy } K = \frac{1}{2}mv^2 = \frac{h^2}{2m\lambda^2}$
Cut-off wavelength $\lambda_0 = \frac{hc}{eV} = \frac{hc}{K} = \frac{hc}{h^2 / (2m\lambda^2)} = \frac{2mc\lambda^2}{h}$
3. Which one of the following statements is wrong in the context of X-rays generated from a X-ray tube? [JEE 2008 P-I]
a. Wavelength of characteristic X-rays decreases when the atomic number of the target increases.
b. Cut-off wavelength of the continuous X-rays depends on the atomic number of the target
c. Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube
d. Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube
Answer: (b)
Solution:
Cut-off wavelength $\lambda_{\text{cut-off}} = \frac{hc}{eV}$ (independent of atomic number)
4. Photoelectric effect experiments are performed using three different metal plates $p$, $q$ and $r$ having work functions $\phi_p = 2.0\text{ eV}$, $\phi_q = 2.5\text{ eV}$ and $\phi_r = 3.0\text{ eV}$ respectively. A light beam containing wavelengths of $550\text{ nm}$, $450\text{ nm}$ and $350\text{ nm}$ with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is (Take $hc = 1240\text{ eV nm}$). [JEE 2009 P-II]
a.
b.
c.
d.
Answer: (a)
Solution:
$V_0 = \frac{1}{e}\left[\frac{hc}{\lambda} - \phi\right]$
$V_p = \frac{1}{e}\left[\frac{1240}{550} - 2\right] = 0.2545\text{ V}$
$V_q = \frac{1}{e}\left[\frac{1240}{450} - 2.5\text{ eV}\right] = 0.255\text{ V}$
$V_r = \frac{1}{e}\left[\frac{1240}{350} - 3\right] = 0.5428\text{ V}$
If $n$ is the number of photons in unit time then $I = \frac{nhc}{\lambda} \implies I \propto \frac{1}{\lambda}$
5. Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields $\vec{E} = E_0\hat{j}$ and $\vec{B} = B_0\hat{j}$. At time $t = 0$, this charge has velocity $\vec{v}$ in the $x\text{--}y$ plane, making an angle $\theta$ with the x-axis. Which of the following option(s) is (are) correct for time $t > 0$? [JEE 2012 P-I]
a. If $\theta = 0^\circ$, the charge moves in a circular path in the $x\text{--}z$ plane.
b. If $\theta = 0^\circ$, the charge undergoes helical motion with constant pitch along the y-axis
c. If $\theta \neq 0^\circ$, the charge undergoes helical motion with its pitch increasing with time, along the y-axis
d. If $\theta = 90^\circ$, the charge undergoes linear but accelerated motion along the y-axis.
Answer: (c, d)
Solution:
For any value $\theta \neq 90^\circ$ Particle will move along helical path with increasing pitch (because of acceleration along $y$ due to $E$). If $\theta = 90^\circ \implies$ motion will be along a straight line and accelerated due to $E$ (as magnetic force is zero).
6. The anode voltage of a photocell is kept fixed. The wavelength $\lambda$ of the light falling on the cathode is gradually changed. The plate current $I$ of the photocell varies as follows [JEE Main 2013]
a.
b.
c.
d.
Answer: (d)
Solution:
As $\lambda$ will increase $K_{\text{max}}$ will decrease so current should decrease and finally fall to zero when $\lambda_0$ is achieved.
7. The magnetic field in a travelling electromagnetic wave has a peak value of $20\text{ nT}$. The peak value of electric field strength is [JEE Main 2013]
a. $3\text{ V/m}$
b. $6\text{ V/m}$
c. $9\text{ V/m}$
d. $12\text{ V/m}$
Answer: (b)
Solution:
$E_0 = c B_0 = 3 \times 10^8 \times 20 \times 10^{-9} = 6\text{ V/m}$
8. A metal surface is illuminated by light of two different wavelengths $248\text{ nm}$ and $310\text{ nm}$. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_2$ respectively. If the ratio $u_1 : u_2 = 2:1$ and $hc = 1240\text{ eV nm}$, the work function of the metal is nearly [JEE Adv. 2014 P-II]
a. 3.7 eV
b. 3.2 eV
c. 2.8 eV
d. 2.5 eV
Answer: (a)
9. During the propagation of electromagnetic waves in a medium: [JEE Main 2014]
a. Electric energy density is equal to the magnetic energy density.
b. Both electric and magnetic energy densities are zero.
c. Electric energy density is double of the magnetic energy density.
d. Electric energy density is half of the magnetic density.
Answer: (a)
Solution:
Energy of the wave is equally distributed in two fields.
10. For photoelectric effect with incident photon wavelength $\lambda$, the stopping potential is $V_0$. Identify the correct variation(s) of $V_0$ with $\lambda$ and $1/\lambda$. [JEE Adv. 2015 P-I]
a.
b.
c.
d.
Answer: (a, c)
Solution:
For photoelectric emission, $eV_0 = \frac{hc}{\lambda} - \phi \implies V_0 = \frac{hc}{e}\left(\frac{1}{\lambda}\right) - \frac{\phi}{e}$
11. Radiation of wavelength $\lambda$, is incident on a photocell. The fastest emitted electron has speed $u$. If the wavelength is changed to $\frac{3}{4}\lambda$, the speed of the fastest emitted electron will be: [JEE Main 2016]
a. $u' > \left(\frac{4}{3}\right)^{1/2} u$
b. $u' < \left(\frac{4}{3}\right)^{1/2} u$
c. $u' = \left(\frac{4}{3}\right)^{1/2} u$
d. $u' = \left(\frac{3}{4}\right)^{1/2} u$
Answer: (a)
Solution:
$$\frac{hc}{\lambda} = W + \frac{1}{2}mu^2 \text{ ---(i)}$$ $$\frac{4hc}{3\lambda} = W + \frac{1}{2}mu'^2 \text{ ---(ii)}$$12. In a historical experiment to determine Planck’s constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength ($\lambda$) of incident light and the corresponding stopping potential ($V_0$) are given below:
$\lambda (\mu\text{m})$ | $V_0 (\text{Volt})$
0.3 | 2.0
0.4 | 1.0
0.5 | 0.4
Given that $c = 3 \times 10^8\text{ m s}^{-1}$ and $e = 1.6 \times 10^{-19}\text{ C}$, Planck’s constant (in units of $\text{J s}$) found from such an experiment is [JEE Adv. 2016 P-I]
a. $6.0 \times 10^{-34}$
b. $6.4 \times 10^{-34}$
c. $6.6 \times 10^{-34}$
d. $6.8 \times 10^{-34}$
Answer: (b)
Solution:
$$\frac{hc}{\lambda} - \phi = eV_0$$Subtracting the equations for $\lambda_1 = 0.3\,\mu\text{m}$ and $\lambda_2 = 0.4\,\mu\text{m}$ gives $h \approx 6.4 \times 10^{-34}\text{ J-s}$.
13. Arrange the following electromagnetic radiations per quantum in the order of increasing energy: [JEE Main 2016]
A : Blue light
B : Yellow light
C : X-ray
D : Radio wave.
a. D, B, A, C
b. A, B, D, C
c. C, A, B, D
d. B, A, D, C
Answer: (a)
Solution:
Energy order: Radio wave < Yellow light < Blue light < X-ray.
14. An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the bulb is powered at constant voltage, which of the following statement(s) is (are) true? [JEE Adv. 2016 P-I]
a. The temperature distribution over the filament is uniform
b. The resistance over small sections of the filament decreases with time
c. The filament emits more light at higher band of frequencies before it breaks up
d. The filament consumes less electrical power towards the end of the life of the bulb
Answer: (c, d)
Solution:
Towards the end of the life, the filament will become thinner. Resistance will increase and so consumed power will be less, so it will emit less light. Temperature distribution will be non-uniform. At the position where temperature is maximum, filament will break.
15. Light of wavelength $\lambda_{\text{ph}}$ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is $\phi$ and the anode is a wire mesh of conducting material kept at a distance $d$ from the cathode. A potential difference $V$ is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is $\lambda_e$, which of the following statement(s) is (are) true? [JEE Adv. 2016 P-II]
a. $\lambda_e$ increases at the same rate as $\lambda_{\text{ph}}$ for $\lambda_{\text{ph}} < hc/\phi$
b. For large potential difference ($eV \gg \phi$), $\lambda_e$ is approximately halved if $V$ is made four times.
c. $\lambda_e$ is approximately halved, if $d$ is doubled
d. $\lambda_e$ decreases with increase in $\phi$ and $\lambda_{\text{ph}}$
Answer: (b)
Solution:
For large potential difference, energy of electron $E \approx eV$.
$\lambda_e = \frac{h}{\sqrt{2mE}} = \frac{h}{\sqrt{2meV}} \implies \lambda_e \propto \frac{1}{\sqrt{V}}$
If $V$ is made 4 times, $\lambda_e$ is halved.
16. An electron beam is accelerated by a potential difference $V$ to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If $\lambda_{\text{min}}$ is the smallest possible wavelength of X-ray in the spectrum, the variation of $\log \lambda_{\text{min}}$ with $\log V$ is correctly represented in: [JEE Main 2017]
a.
b.
c.
d.
Answer: (d)
Solution:
$\lambda_{\text{min}} = \frac{hc}{eV} \implies \log(\lambda_{\text{min}}) = \log(hc) - \log(e) - \log(V)$
This is a straight line with a negative slope.
17. A particle of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $m/2$ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths $\lambda_A$ to $\lambda_B$ after the collision is: [JEE Main 2017]
a. $\lambda_A = 2\lambda_B$
b. $\lambda_A = \frac{2}{3}\lambda_B$
c. $\lambda_A = \frac{1}{2}\lambda_B$
d. $\lambda_A = \frac{1}{3}\lambda_B$
Answer: (a)
18. An EM wave from air enters a medium. The electric fields are $\vec{E}_1 = E_{01}\hat{i}\cos\left[\frac{2\pi}{\lambda}(z - ct)\right]$ in air and $\vec{E}_2 = E_{02}\hat{i}\cos[k(z - vt)]$ in medium, where the wave number $k$ and frequency $\nu$ refer to their values in air. The medium is non-magnetic. If $\varepsilon_{r1}$ and $\varepsilon_{r2}$ refer to relative permittivities for air and medium respectively, which of the following options is correct: [JEE Main 2018]
a. $\frac{\varepsilon_{r1}}{\varepsilon_{r2}} = 1$
b. $\frac{\varepsilon_{r1}}{\varepsilon_{r2}} = 4$
c. $\frac{\varepsilon_{r1}}{\varepsilon_{r2}} = 2$
d. $\frac{\varepsilon_{r1}}{\varepsilon_{r2}} = \frac{1}{4}$
Answer: (d)
19. An electron from various excited states of hydrogen atom emits radiation to come to the ground state. Let $\lambda_n, \lambda_g$ be the de Broglie wavelength of the electron in the $n^{\text{th}}$ state and the ground state respectively. Let $\Lambda_n$ be the wavelength of the emitted photon in the transition from the $n^{\text{th}}$ state to the ground state. For large $n$, $\Lambda_n \approx A\lambda_n^2 + B$ (A, B are constants). [JEE Main 2018]
a. $\Lambda_n \approx 2\lambda_n^2$
b. $\Lambda_n \approx \frac{m c}{h}\lambda_n^2 + A$
c. $\Lambda_n \approx A\lambda_n + B$
d. $\Lambda_n \approx A\lambda_n^2 + B$
Answer: (b)
20. On interchanging the resistances, the balance point of a meter bridge shifts to the left by $10\text{ cm}$. The resistance of their series combination is $1\text{ k}\Omega$. How much was the resistance on the left slot before interchanging the resistances? [JEE Main 2018]
a. $910\ \Omega$
b. $990\ \Omega$
c. $505\ \Omega$
d. $550\ \Omega$
Answer: (d)
21. The magnetic field of an electromagnetic wave is given by: $\vec{B} = 1.6 \times 10^{-6}\cos(2\pi \times 10^6 z + 3\pi \times 10^{15} t)(\hat{i} + 2\hat{j})\text{ Wb/m}^2$. The associated electric field will be: [JEE Main 2019]
a. $\vec{E} = -4.8 \times 10^2 \cos(2\pi \times 10^6 z + 3\pi \times 10^{15} t)(2\hat{i} - \hat{j})\text{ V/m}$
b. $\vec{E} = 4.8 \times 10^2 \cos(2\pi \times 10^6 z + 3\pi \times 10^{15} t)(2\hat{i} + \hat{j})\text{ V/m}$
c. $\vec{E} = -4.8 \times 10^2 \cos(2\pi \times 10^6 z + 3\pi \times 10^{15} t)(\hat{i} - 2\hat{j})\text{ V/m}$
d. $\vec{E} = 4.8 \times 10^2 \cos(2\pi \times 10^6 z + 3\pi \times 10^{15} t)(-2\hat{i} + \hat{j})\text{ V/m}$
Answer: (d)
22. Two particle are moving perpendicular to each other with de-Broglie wave length $\lambda_1$ and $\lambda_2$. If they collide and stick then the de-Broglie wave length of system after collision is: [JEE Main 2019]
a. $\lambda = \frac{\lambda_1\lambda_2}{\sqrt{\lambda_1^2 + \lambda_2^2}}$
b. $\lambda = \frac{\lambda_1\lambda_2}{\lambda_1 + \lambda_2}$
c. $\lambda = \sqrt{\lambda_1^2 + \lambda_2^2}$
d. $\lambda = \lambda_1 + \lambda_2$
Answer: (a)
23. The energy associated with electric field is ($U_E$) and with magnetic field is ($U_B$) for an electromagnetic wave in free space. Then: [JEE Main 2019]
a. $U_E = U_B^2$
b. $U_E < U_B$
c. $U_E = U_B$
d. $U_E > U_B$
Answer: (c)
Solution:
Energy density of electric and magnetic fields in an electromagnetic wave are equal: $U_E = U_B$.
24. The magnetic field associated with a light wave is given, at the origin, by $B = B_0 [\sin(3.14 \times 10^7 ct) + \sin(6.28 \times 10^7 ct)]$. If this light falls on a silver plate having a work function of $4.7\text{ eV}$, what will be the maximum kinetic energy of the photo electrons? ($c = 3 \times 10^8\text{ ms}^{-1}$, $h = 6.6 \times 10^{-34}\text{ J-s}$) [JEE Main 2019]
a. 7.72 eV
b. 8.52 eV
c. 12.5 eV
d. 6.82 eV
Answer: (a)
Solution:
Using the higher frequency component to find maximum kinetic energy: $KE_{\text{max}} = \text{E}_{\text{photon}} - \phi \approx 7.675\text{ eV} \approx 7.72\text{ eV}$.
25. A plane electromagnetic wave of frequency $50 \times 10^6\text{ MHz}$ travels in free space along the positive x-direction. At a particular point in space and time, $\vec{E} = 6.3\hat{j}\text{ V/m}$. The corresponding magnetic field $\vec{B}$ at that point will be: [JEE Main 2019]
a. $18.9 \times 10^{-8}\hat{k}\text{ T}$
b. $6.3 \times 10^{-8}\hat{k}\text{ T}$
c. $2.1 \times 10^{-8}\hat{k}\text{ T}$
d. $18.9 \times 10^{-8}\hat{j}\text{ T}$
Answer: (c)
Solution:
$B = \frac{E}{c} = \frac{6.3}{3 \times 10^8} = 2.1 \times 10^{-8}\text{ T}$ along $\hat{k}$.
26. Surface of certain metal is first illuminated with light of wavelength $\lambda_1 = 350\text{ nm}$ and then, by light of wavelength $\lambda_2 = 540\text{ nm}$. It is found that the maximum speed of the photo electrons in the two case differ by a factor of 2. The work function of the metal (in eV) is close to: (Energy of photon $=\frac{1240}{\lambda \text{ (in nm)}}\text{ eV}$) [JEE Main 2019]
a. 1.8
b. 1.4
c. 2.5
d. 5.6
Answer: (a)
Solution:
$$\frac{hc}{\lambda_1} = \phi + \frac{1}{2} m (2v)^2$$ $$\frac{hc}{\lambda_2} = \phi + \frac{1}{2} m v^2$$ $$\frac{\frac{hc}{\lambda_1} - \phi}{\frac{hc}{\lambda_2} - \phi} = 4$$ $$\frac{hc}{\lambda_1} - \phi = 4 \frac{hc}{\lambda_2} - 4\phi$$ $$3\phi = \frac{4hc}{\lambda_2} - \frac{hc}{\lambda_1}$$ $$\phi = \frac{1}{3} hc \left( \frac{4}{\lambda_2} - \frac{1}{\lambda_1} \right)$$ $$\phi = \frac{1}{3} \times 1240 \left( \frac{4}{540} - \frac{1}{350} \right) = 1.8\text{ eV}$$27. The electric field of a plane polarised electromagnetic wave in free space at time $t = 0$ is given by an expression $\vec{E}(x, y) = 10 \hat{j} \cos[6x + 8z]$. The magnetic field $\vec{B}(x, z, t)$ is given by: ($c$ is the velocity of light) [JEE Main 2019]
a. $\frac{1}{c} (6\hat{k} + 8\hat{i}) \cos[(6x + 8z - 10^8 c t)]$
b. $\frac{1}{c} (6\hat{k} - 8\hat{i}) \cos[(6x + 8z - 10^8 c t)]$
c. $\frac{1}{c} (6\hat{k} + 8\hat{i}) \cos[(6x + 8z + 10^8 c t)]$
d. $\frac{1}{c} (6\hat{k} - 8\hat{i}) \cos[(6x + 8z + 10^8 c t)]$
Answer: (b)
Solution:
$\vec{E} = 10 \hat{j} \cos[(6\hat{i} + 8\hat{k}) \cdot (x\hat{i} + z\hat{k})] = 10 \cos[\vec{K} \cdot \vec{r}] \hat{j}$
$\therefore \vec{K} = 6\hat{i} + 8\hat{k}$; direction of waves travel i.e direction of ‘c’.
$\therefore$ Direction of $\hat{B}$ will be along $\hat{C} \times \hat{E} = \frac{6\hat{i} + 8\hat{k}}{10} \times \hat{j} = \frac{6\hat{k} - 8\hat{i}}{10}$
Mag. of $\vec{B}$ will be $\frac{E}{c} = \frac{10}{c}$
$\therefore \vec{B} = \frac{10}{c} \left(\frac{6\hat{k} - 8\hat{i}}{10}\right) = \frac{1}{c}(6\hat{k} - 8\hat{i})\cos[(6x + 8z - 10^8 c t)]$
28. A metal plate of area $1 \times 10^{-4}\text{ m}^2$ is illuminated by a radiation of intensity $16\text{ mW/m}^2$. The work function of the metal is $5\text{ eV}$. The energy of the incident photons is $10\text{ eV}$ and only $10\%$ of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be: [$1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$] [JEE Main 2019]
a. $10^{10}$ and $5\text{ eV}$
b. $10^{14}$ and $10\text{ eV}$
c. $10^{12}$ and $5\text{ eV}$
d. $10^{11}$ and $5\text{ eV}$
Answer: (d)
Solution:
$E = n_{\text{Photon}} \times E_{\text{photon}}$
$\frac{n_{\text{Photon}}}{t} = \frac{I A}{E_{\text{photon}}} = \frac{16 \times 10^{-3} \times 1 \times 10^{-4}}{10 \times 1.6 \times 10^{-19}} = 10^{12}$
Number of photoelectrons emitted per second $= 10\% \text{ of } 10^{12} = 10^{11}$
Maximum kinetic energy $= E - \phi = 10\text{ eV} - 5\text{ eV} = 5\text{ eV}$
29. If the magnetic field of a plane electromagnetic wave is given by: (The speed of light $= 3 \times 10^8\text{ m/s}$)
$$B = 100 \times 10^{-6} \sin \left[ 2 \pi \times 15 \times 10^6 \left( x - c t \right) \right]$$then the maximum electric field associated with it is: [JEE Main 2019]
a. $4 \times 10^4\text{ N/C}$
b. $4.5 \times 10^4\text{ N/C}$
c. $6 \times 10^4\text{ N/C}$
d. $3 \times 10^4\text{ N/C}$
Answer: (d)
Solution:
$E_0 = B_0 \times c = 100 \times 10^{-6} \times 3 \times 10^8 = 3 \times 10^4\text{ N/C}$
$\therefore$ Correct answer is $3 \times 10^4\text{ N/C}$
30. In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5 \times 10^{-12}\text{ m}$, the minimum electron energy required is close to: [JEE Main 2019]
a. $100\text{ keV}$
b. $500\text{ keV}$
c. $25\text{ keV}$
d. $1\text{ keV}$
Answer: (c)
Solution:
$\lambda = \frac{h}{p}$ where $\{\lambda = 7.5 \times 10^{-12}\text{ m}\}$
$p = \frac{h}{\lambda}$
$\text{KE} = \frac{p^2}{2m} = \frac{(h/\lambda)^2}{2m}$
$$\text{KE} = \frac{\left( \frac{6.6 \times 10^{-34}}{7.5 \times 10^{-12}} \right)^2}{2 \times 9.1 \times 10^{-31}}\text{ J}$$$\text{KE} = 25\text{ keV}$
31. A $27\text{ mW}$ laser beam has a cross-sectional area of $10\text{ mm}^2$. The magnitude of the maximum electric field in this electromagnetic wave is given by [Given permittivity of space $\epsilon_0 = 9 \times 10^{-12}\text{ SI units}$, Speed of light $c = 3 \times 10^8\text{ m/s}$]: [JEE Main 2019]
a. $1\text{ kV/m}$
b. $2\text{ kV/m}$
c. $1.4\text{ kV/m}$
d. $0.7\text{ kV/m}$
Answer: (c)
Solution:
Intensity of EM wave is given by
$$\text{Intensity } I = \frac{\text{Power}}{\text{Area}} = \frac{1}{2} \epsilon_0 E_0^2 c$$ $$\frac{27 \times 10^{-3}}{10 \times 10^{-6}} = \frac{1}{2} \times 9 \times 10^{-12} \times E_0^2 \times 3 \times 10^8$$ $$E_0^2 = 2 \times 10^6 \implies E_0 = \sqrt{2} \times 10^3 = 1.4\text{ kV/m}$$32. In a photoelectric experiment, the wavelength of the light incident on a metal is changed from $300\text{ nm}$ to $400\text{ nm}$. The decrease in the stopping potential is close to: ($\frac{hc}{e} = 1240\text{ nm V}$) [JEE Main 2019]
a. $0.5\text{ V}$
b. $1.0\text{ V}$
c. $1.5\text{ V}$
d. $2.0\text{ V}$
Answer: (b)
Solution:
$$\frac{hc}{\lambda_1} = \phi + eV_1 \quad \dots(i)$$ $$\frac{hc}{\lambda_2} = \phi + eV_2 \quad \dots(ii)$$(i) – (ii)
$$e(V_1 - V_2) = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right)$$ $$V_1 - V_2 = \left(\frac{hc}{e}\right) \left( \frac{\lambda_2 - \lambda_1}{\lambda_1 \lambda_2} \right) = 1240\text{ nm V} \times \frac{100\text{ nm}}{300\text{ nm} \times 400\text{ nm}} = 1\text{ V}$$33. A particle of mass $m$ and charge $q$ is in an electric and magnetic field given by $\vec{E} = 2\hat{i} + 3\hat{j}; \vec{B} = 4\hat{j} + 6\hat{k}$. The charged particle is shifted from the origin to the point $P(x = 1 ; y = 1)$ along a straight path. The magnitude of the total work done is: [JEE Main 2019]
a. $(0.35)q$
b. $(0.15)q$
c. $(2.5)q$
d. $5q$
Answer: (d)
Solution:
$$\vec{F}_{\text{net}} = q(\vec{E} + \vec{v} \times \vec{B}) = q(2\hat{i} + 3\hat{j}) + q(\vec{v} \times \vec{B})$$ $$W = \vec{F}_{\text{net}} \cdot \vec{S} = q(2\hat{i} + 3\hat{j}) \cdot (\hat{i} + \hat{j}) = q(2 + 3) = 5q$$34. An electromagnetic wave of intensity $50\text{ Wm}^{-2}$ enters in a medium of refractive index '$n$' without any loss. The ratio of the magnitudes of electric fields, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by: [JEE Main 2019]
a. $\left( \frac{1}{n}, n \right)$
b. $\left( \frac{1}{\sqrt{n}}, \sqrt{n} \right)$
c. $(\sqrt{n}, \sqrt{n})$
d. $\left( \frac{1}{\sqrt{n}}, \frac{1}{\sqrt{n}} \right)$
Answer: (b)
Solution:
$$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$ $$v = \frac{1}{\sqrt{\mu k \epsilon_0}}$$[For transparent medium $\mu_r \approx 1$]
$$\therefore \frac{c}{v} = \sqrt{k} = n$$$\frac{1}{2} \epsilon_0 E_0^2 c = \text{intensity} = \frac{1}{2} \epsilon_0 k E^2 v$
$$\therefore E_0^2 c = k E^2 v$$ $$\implies \frac{E_0^2}{E^2} = \frac{k v}{c} = \frac{n^2}{n} = n \implies \frac{E_0}{E} = \sqrt{n}$$Similarly,
$$\frac{B_0^2}{\mu_0} c = \frac{B^2}{\mu_0} v \implies \frac{B_0}{B} = \frac{1}{\sqrt{n}}$$35. If the de Broglie wave length of an electron is equal to $10^{-3}$ times the wavelength of a photon of frequency $6 \times 10^{14}\text{ Hz}$, then the speed of electron is equal to: (Speed of light $= 3 \times 10^8\text{ m/s}$, Planck's constant $= 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$, Mass of electron $= 9.1 \times 10^{-31}\text{ kg}$) [JEE Main 2019]
a. $1.45 \times 10^6\text{ m/s}$
b. $1.7 \times 10^6\text{ m/s}$
c. $1.8 \times 10^6\text{ m/s}$
d. $1.1 \times 10^6\text{ m/s}$
Answer: (a)
Solution:
$$\frac{h}{m v} = 10^{-3} \left( \frac{c}{\nu} \right) = 10^{-3} \left( \frac{3 \times 10^8}{6 \times 10^{14}} \right)$$ $$\frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times v} = \frac{3 \times 10^8 \times 10^{-3}}{6 \times 10^{14}}$$ $$v = 1.45 \times 10^6\text{ m/s}$$36. The mean intensity of radiation on the surface of the Sun is about $10^8\text{ W/m}^2$. The rms value of the corresponding magnetic field is closest to: [JEE Main 2019]
a. $10^2\text{ T}$
b. $10^{-4}\text{ T}$
c. $1\text{ T}$
d. $10^{-2}\text{ T}$
Answer: (b)
Solution:
$I = \epsilon_0 c E_{\text{rms}}^2$ and $E_{\text{rms}} = c B_{\text{rms}}$
$I = \epsilon_0 c^3 B_{\text{rms}}^2 = \frac{1}{\mu_0} c B_{\text{rms}}^2$
$$B_{\text{rms}} \approx 10^{-4}\text{ T}$$37. In a Franck-Hertz experiment, an electron of energy $5.6\text{ eV}$ passes through mercury vapour and emerges with an energy $0.7\text{ eV}$. The minimum wavelength of photons emitted by mercury atoms is close to: [JEE Main 2019]
a. $2020\text{ nm}$
b. $220\text{ nm}$
c. $250\text{ nm}$
d. $1700\text{ nm}$
Answer: (c)
Solution:
$$\lambda = \frac{1240}{5.6 - 0.7}\text{ nm} = \frac{1240}{4.9}\text{ nm} \approx 250\text{ nm}$$38. A particle A of mass '$m$' and charge '$q$' is accelerated by a potential difference of $50\text{ V}$. Another particle B of mass '$4m$' and charge '$q$' is accelerated by a potential difference of $2500\text{ V}$. The ratio of de-Broglie wavelengths $\frac{\lambda_A}{\lambda_B}$ is close to: [JEE Main 2019]
a. 10.00
b. 14.14
c. 4.47
d. 0.07
Answer: (b)
Solution:
K.E. acquired by charge $= K = qV$
$$\lambda = \frac{h}{P} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}$$ $$\therefore \frac{\lambda_A}{\lambda_B} = \sqrt{\frac{m_B q_B V_B}{m_A q_A V_A}} = \sqrt{\frac{4m \cdot q \cdot 2500}{m \cdot q \cdot 50}} = \sqrt{200} = 10\sqrt{2} = 14.14$$Matrix Match/Column Match
39. Match the statement of Column I with those in Column II: [JEE Main 2015]
Column I
(A) Franck-Hertz Experiment
(B) Photoelectric Experiment
(C) Davisson–Germer Experiment
Column II
1. Particle nature of light
2. Discrete energy levels of atom
3. Wave nature of electron
4. Structure of atom
a. A$\rightarrow$2; B$\rightarrow$1; C$\rightarrow$3
b. A$\rightarrow$3; B$\rightarrow$2; C$\rightarrow$4
c. A$\rightarrow$1; B$\rightarrow$3; C$\rightarrow$2
d. A$\rightarrow$3; B$\rightarrow$1; C$\rightarrow$2
Answer: (a)
Solution:
(A) Franck-Hertz experiment $\rightarrow$ Discrete energy levels of atom.
(B) Photoelectric experiment $\rightarrow$ Particle nature of light
(C) Davisson–Germer Experiment $\rightarrow$ Wave nature of electron.
Statement or Assertion and Reason
40. Statement-I: A vertical iron rod has a coil of wire wound over it at the bottom end. An alternating current flows in the coil. The rod goes through a conducting ring as shown in the figure. The ring can float at a certain height above the coil. Statement-II: In the above situation, a current is induced in the ring which interacts with the horizontal component of the magnetic field to produce an average force in the upward direction. [JEE 2007 P-II]
a. Statement-I is True, Statement-II is True; Statement-II is a correct explanation for Statement-I.
b. Statement-I is True, Statement-II is True; Statement-II is not a correct explanation for Statement-I.
c. Statement-I is True, Statement-II is False.
d. Statement-I is False, Statement-II is True.
Answer: (a)
Solution:
The induced current in the ring creates a magnetic dipole moment that interacts with the diverging magnetic field (specifically its horizontal component) of the coil, resulting in a net upward magnetic force that balances gravity, allowing the ring to float.
Paragraphs
Paragraph for Question Nos. 41 to 43
When a particle is restricted to move along x-axis between $x = 0$ and $x = a$, where $a$ is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends $x = 0$ and $x = a$. The wavelength of this standing wave is related to the linear momentum $p$ of the particle according to the de Broglie relation. The energy of the particle of mass $m$ is related to its linear momentum as $E = p^2/2m$. Thus, the energy of the particle can be denoted by a quantum number ‘$n$’ taking values $1, 2, 3, \dots$ ($n = 1$, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving in the line $x = 0$ to $x = a$. Take $h = 6.6 \times 10^{-34}\text{ J s}$ and $e = 1.6 \times 10^{-19}\text{ C}$. [JEE 2009 P-I]
41. The allowed energy for the particle for a particular value of $n$ is proportional to
a. $a^{-2}$
b. $a^{-3/2}$
c. $a^{-1}$
d. $a^2$
Answer: (a)
Solution:
$$\frac{n\lambda}{2} = a \implies \lambda = \frac{2a}{n}$$ $$\text{de Broglie: } \lambda = \frac{h}{p} \implies \frac{2a}{n} = \frac{h}{p} \implies p = \frac{nh}{2a}$$ $$E = \frac{p^2}{2m} = \frac{n^2 h^2}{8m a^2}$$ $$\implies E \propto a^{-2}$$42. If the mass of the particle is $m = 1.0 \times 10^{-30}\text{ kg}$ and $a = 6.6\text{ nm}$, the energy of the particle in its ground state is closest to
a. $0.8\text{ meV}$
b. $8\text{ meV}$
c. $80\text{ meV}$
d. $800\text{ meV}$
Answer: (b)
Solution:
$$E = \frac{h^2}{8 a^2 m} = \frac{(6.6 \times 10^{-34})^2}{8 \times (6.6 \times 10^{-9})^2 \times 10^{-30} \times 1.6 \times 10^{-19}}\text{ eV} = 8\text{ meV}$$43. The speed of the particle that can take discrete values is proportional to
a. $n^{-3/2}$
b. $n^{-1}$
c. $n^{1/2}$
d. $n$
Answer: (d)
Solution:
$$p = \frac{nh}{2a} \implies m v = \frac{nh}{2a} \implies v = \frac{nh}{2am} \implies v \propto n$$Paragraph for Question No. 44
In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the question below, $[E]$ and $[B]$ stand for dimensions of electric and magnetic fields respectively, while $[\epsilon_0]$ and $[\mu_0]$ stand for dimensions of the permittivity and permeability of free space respectively. $[L]$ and $[T]$ are dimensions of length and time respectively. All the quantities are given in SI units. [JEE Adv. 2018 P-I]
44. The relation between $[E]$ and $[B]$ is
a. $[E] = [B][L][T]$
b. $[E] = [B][L]^{-1}[T]$
c. $[E] = [B][L][T]^{-1}$
d. $[E] = [B][L]^{-1}[T]^{-1}$
Answer: (c)
Solution:
$$v = \frac{E}{B} \implies [E] = [B] [v] = [B] [L] [T]^{-1}$$ $$[E] = [B][L][T]^{-1}$$Integer and Subjective
45. The potential energy of a particle of mass $m$ is given by $V(x) = E_0$ for $0 \le x \le 1$ and $V(x) = 0$ for $x > 1$. $\lambda_1$ and $\lambda_2$ are the de-Broglie wavelengths of the particle, when $0 \le x \le 1$ and $x > 1$ respectively. If the total energy of particle is $2E_0$, find $\lambda_1 / \lambda_2$. [JEE 2005]
Solution:
K.E. $= 2E_0 - E_0 = E_0$ (for $0 \le x \le 1$)
$$\implies \lambda_1 = \frac{h}{\sqrt{2mE_0}}$$K.E. $= 2E_0 - 0 = 2E_0$ (for $x > 1$)
$$\implies \lambda_2 = \frac{h}{\sqrt{4mE_0}}$$ $$\implies \frac{\lambda_1}{\lambda_2} = \sqrt{2}$$46. What will be the minimum angle of incidence such that the total internal reflection occurs on both the surfaces? [JEE 2005]
Solution:
For first surface:
$$\mu_1 \sin c_1 = \mu_2 \sin 90^\circ \implies \sqrt{2} \sin c_1 = \sqrt{3} \sin 90^\circ \text{ (not possible directly, } c_1 = 45^\circ \text{ for boundary context)}$$For second surface:
$$\mu_2 \sin c_2 = \mu_3 \sin 90^\circ \implies \sqrt{3} \sin c_2 = 2 \sin 90^\circ$$ $$\implies c_2 = 60^\circ$$$\therefore$ Minimum angle of incidence $= \text{Max}\{c_1, c_2\} = 60^\circ$
47. An $\alpha$-particle and a proton are accelerated from rest by a potential difference of $100\text{ V}$. After this, their de Broglie wavelengths are $\lambda_\alpha$ and $\lambda_p$ respectively. The ratio $\frac{\lambda_p}{\lambda_\alpha}$ to the nearest integer, is: [JEE 2010 P-I]
Answer: 3
Solution:
$$\frac{1}{2} m v^2 = qV$$ $$\lambda = \frac{h}{m v} = \frac{h}{\sqrt{2mqV}}$$ $$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{\frac{4m_p \cdot 2e}{m_p \cdot e}} = \sqrt{8} \approx 2.83 \approx 3$$48. A silver sphere of radius $1\text{ cm}$ and work function $4.7\text{ eV}$ is suspended from an insulating thread in free space. It is under continuous illumination of $200\text{ nm}$ wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is $A \times 10^Z$ (where $1 < A < 10$). The value of '$Z$' is: [JEE 2011 P-II]
Answer: 7
Solution:
Stopping potential $V = \frac{hc}{\lambda e} - \phi = 6.2\text{ eV} - 4.7\text{ eV} = 1.5\text{ eV}$
$$V = \frac{K q}{r} \implies 1.5 = \frac{9 \times 10^9 \times n \times 1.6 \times 10^{-19}}{10^{-2}}$$ $$n = \frac{1.5 \times 10^{-2}}{9 \times 10^9 \times 1.6 \times 10^{-19}} = 1.04 \times 10^7$$ $$\implies Z = 7$$49. A proton is fired from very far away towards a nucleus with charge $Q = 120e$, where $e$ is the electronic charge. It makes a closest approach of $10\text{ fm}$ to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is: (take the proton mass, $m_p = (5/3) \times 10^{-27}\text{ kg}$; $h/e = 4.2 \times 10^{-15}\text{ J s / C}$; $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\text{ N m}^2/\text{C}^2$; $1\text{ fm} = 10^{-15}\text{ m}$) [JEE 2012 P-I]
Answer: 7
Solution:
By Conservation of energy:
$$\frac{1}{2} m v_\infty^2 = \frac{1}{4\pi\epsilon_0} \frac{e(120e)}{10\text{ fm}}$$Assuming the nucleus to be considerably massive, we can disregard its motion.
Let momentum of proton be $p = m v_\infty$
$$\frac{p^2}{2m} = \frac{1}{4\pi\epsilon_0} \frac{120 e^2}{10 \times 10^{-15}}$$ $$p^2 = 2 \times \frac{5}{3} \times 10^{-27} \times \frac{120 \times 9 \times 10^9 e^2}{10^{-14}} = 3600 \times 10^{-42} e^2$$ $$p = 60 \times 10^{-21} e$$ $$\lambda = \frac{h}{p} = \frac{h}{60 \times 10^{-21} e} = \frac{4.2 \times 10^{-15}}{60 \times 10^{-21}} = 7 \times 10^{-15}\text{ m}$$ $$\therefore \lambda = 7\text{ fm}$$50. The work functions of Silver and Sodium are $4.6$ and $2.3\text{ eV}$, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is [JEE Adv. 2013 P-I]
Answer: 1
Solution:
Slope of stopping potential vs frequency graph is $\frac{h}{e} = \text{constant}$.
$$\text{Ratio} = 1$$51. An electron in an excited state of $\text{Li}^{2+}$ ion has angular momentum $3h / 2\pi$. The de Broglie wavelength of the electron in this state is $p\pi a_0$ (where $a_0$ is the Bohr radius). The value of $p$ is [JEE Adv. 2015 P-II]
Solution:
$$mvr = \frac{nh}{2\pi} = \frac{3h}{2\pi}$$de-Broglie Wavelength:
$$\lambda = \frac{h}{mv} = \frac{2\pi r}{3}$$ $$r = \frac{a_0 n^2}{Z} = \frac{a_0 (3)^2}{3} = 3a_0$$ $$\lambda = \frac{2\pi (3a_0)}{3} = 2\pi a_0$$Comparing with $p\pi a_0$, we get $p = 2$.
52. In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force $F = n \times 10^{-4}\text{ N}$ due to the impact of the electrons. The value of $n$ is ______. Mass of the electron $m_e = 9 \times 10^{-31}\text{ kg}$ and $1.0\text{ eV} = 1.6 \times 10^{-19}\text{ J}$. [JEE Adv. 2018 P-II]
Answer: 24
Solution:
$P = 200\text{ J/s}$
No. of photons per second ($N$):
$$N = \frac{200}{6.25 \times 1.6 \times 10^{-19}} = 2 \times 10^{20}$$ $$\Delta p = \sqrt{2m(\text{KE})} = \sqrt{2m(e V)}$$ $$\Delta p = \sqrt{2 \times 9 \times 10^{-31} \times 1.6 \times 10^{-19} \times 500} = 12 \times 10^{-24}$$ $$\therefore F = \Delta p \cdot N = 12 \times 10^{-24} \times 2 \times 10^{20}$$ $$F = 24 \times 10^{-4}\text{ N}$$Comparing with $n \times 10^{-4}\text{ N}$, the value of $n = 24$.
53. Perfectly reflecting mirror of mass M mounted on a spring constitutes a spring mass system of angular frequency $\Omega$ such that $\frac{4\pi M\Omega}{h} = 10^{24}\text{ m}^{-2}$ where h is planck constant. N photons of wavelength $\lambda = 8\pi \times 10^{-6}\text{ m}$ strikes the mirror simultaneously at normal incidence such that the mirror gets displaced by $1\mu\text{m}$. If the value of N is $x \times 10^{12}$ then find value of x. [JEE Adv. 2019 P-II]
Answer: 1
Solution:
Momentum transferred on mirror:
$$p = \frac{2Nh}{\lambda}$$At mean position:
$$\frac{2Nh}{\lambda} = M V$$ $$V = A \Omega \quad \text{(where, } A = 1\mu\text{m)}$$ $$\frac{2Nh}{\lambda} = M A \Omega \quad \text{(where, } \lambda = 8\pi \times 10^{-6}\text{ m)}$$ $$\frac{N 2 h}{8\pi \times 10^{-6}} = M (10^{-6}) \Omega$$ $$\frac{2h N}{8\pi \times 10^{-6}} = M \Omega \times 10^{-6}$$ $$\frac{4\pi M \Omega}{h} \times 10^{-12} = N$$ $$10^{24} \times 10^{-12} = N$$ $$N = 1 \times 10^{12}$$$x = 1$
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