#
Multiple Choice Questions
One and More than One Correct
1. The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is: [JEE 2007 P-I]
a. 802 nm
b. 823 nm
c. 1882 nm
d. 1648 nm
Answer: (b)
Solution:
Transition from $\infty$ to $n = 3$ will produce smallest wavelength in infrared region.
2. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561Ã….The wavelength of the second spectral line in the Balmer series of singly-ionised helium atom is: [JEE 2011 P-I]
a. 1215 Ã…
b. 1640 Ã…
c. 2430 Ã…
d. 4687 Ã…
Answer: (a)
Solution:
$$\frac{1}{\lambda_1} = R \left[\frac{1}{4} - \frac{1}{9}\right] = \frac{5}{36}R$$ $$\frac{1}{\lambda_2} = 4R \left[\frac{1}{4} - \frac{1}{16}\right] = \frac{3}{4}R$$ $$\Rightarrow \frac{\lambda_2}{\lambda_1} = \frac{5}{36} \times \frac{4}{3} = \frac{5}{27}$$ $$\Rightarrow \lambda_2 = \frac{5}{27} \times 6561 = 1215\text{ Ã…}$$3. The radius of the orbit of an electron in a Hydrogen-like atom is $4.5 a_0$ where $a_0$ is the Bohr radius. Its orbital angular momentum is $\frac{3h}{2\pi}$. It is given that h is Planck’s constant and R is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are) [JEE Adv. 2013 P-II]
a. $\frac{9}{32R}$
b. $\frac{9}{16R}$
c. $\frac{9}{5R}$
d. $\frac{4}{3R}$
Answer: (a, c)
Solution:
Given data
$$\frac{n^2}{Z} a_0 = 4.5 a_0 \quad \text{. . .(i)}$$ $$\frac{nh}{2\pi} = \frac{3h}{2\pi} \quad \text{. . .(ii)}$$So, $n = 3$ and $Z = 2$
So, possible wavelengths are
$$\frac{1}{\lambda_1} = R Z^2 \left[\frac{1}{1^2} - \frac{1}{3^2}\right] \Rightarrow \lambda_1 = \frac{9}{32R}$$ $$\frac{1}{\lambda_2} = R Z^2 \left[\frac{1}{1^2} - \frac{1}{2^2}\right] \Rightarrow \lambda_2 = \frac{1}{3R}$$ $$\frac{1}{\lambda_3} = R Z^2 \left[\frac{1}{2^2} - \frac{1}{3^2}\right] \Rightarrow \lambda_3 = \frac{9}{5R}$$4. In a hydrogen like atom electron makes transition from an energy level with quantum number n to another with quantum number (n − 1). If n >> 1, the frequency of radiation emitted is proportional to: [JEE Main 2013]
a. $\frac{1}{n}$
b. $\frac{1}{n^2}$
c. $\frac{1}{n^{2/3}}$
d. $\frac{1}{n^3}$
Answer: (d)
Solution:
$$\frac{1}{\lambda} = R \left(\frac{1}{(n-1)^2} - \frac{1}{n^2}\right)$$ $$\nu = \frac{c}{\lambda} = Rc \left(\frac{n^2 - (n-1)^2}{n^2(n-1)^2}\right)$$ $$\nu = Rc \left(\frac{2n-1}{n^2(n-1)^2}\right) \approx Rc \left(\frac{2n}{n^4}\right)$$ $$\Rightarrow \nu = \frac{2Rc}{n^3}$$ $$\Rightarrow \nu \propto \frac{1}{n^3}$$5. If $\lambda_{\text{Cu}}$ is the wavelength of $K_\alpha$ X-ray line of copper (atomic number 29) and $\lambda_{\text{Mo}}$ is the wavelength of the $K_\alpha$ X-ray line of molybdenum (atomic number 42), then the ratio $\lambda_{\text{Cu}} / \lambda_{\text{Mo}}$ is close to [JEE Adv. 2014 P-II]
a. 1.99
b. 2.14
c. 0.50
d. 0.48
Answer: (b)
Solution:
$$\frac{\lambda_{\text{Cu}}}{\lambda_{\text{Mo}}} = \left(\frac{Z_{\text{Mo}} - 1}{Z_{\text{Cu}} - 1}\right)^2 = \left(\frac{42 - 1}{29 - 1}\right)^2 = \left(\frac{41}{28}\right)^2 \approx 2.14$$6. Hydrogen ($_1\text{H}^1$), Deuterium ($_1\text{H}^2$), singly ionised Helium ($_2\text{He}^4$)$^+$ and doubly ionised lithium ($_3\text{Li}^6$)$^{++}$ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are $\lambda_1, \lambda_2, \lambda_3$ and $\lambda_4$ respectively then approximately which one of the following is correct? [JEE Main 2014]
a. $\lambda_1 = 1 \lambda_2 = 4 \lambda_3 = 9 \lambda_4$
b. $\lambda_1 = 2 \lambda_2 = 3 \lambda_3 = 4 \lambda_4$
c. $4 \lambda_1 = 2 \lambda_2 = 2 \lambda_3 = \lambda_4$
d. $\lambda_1 = 2 \lambda_2 = 2 \lambda_3 = 4 \lambda_4$
Answer: (a)
Solution:
$$\frac{1}{\lambda} = R Z^2 \left(\frac{1}{1^2} - \frac{1}{2^2}\right) \cdot \frac{1}{1 + \frac{m_e}{M}}$$ $$\lambda_0 = \frac{4}{3 Z^2 R} \cdot \left(1 + \frac{m_e}{M}\right)$$[$M$ = mass of nucleus]
$$\lambda_1 \propto \frac{1}{1^2} \left(1 + \frac{m_e}{M_p}\right)$$ $$\lambda_2 \propto \frac{1}{1^2} \left(1 + \frac{m_e}{2M_p}\right)$$ $$\lambda_3 \propto \frac{1}{2^2} \left(1 + \frac{m_e}{4M_p}\right)$$ $$\lambda_4 \propto \frac{1}{3^2} \left(1 + \frac{m_e}{6M_p}\right)$$If $\frac{m_e}{M_p} \ll 1$, we have $\lambda_1 = \lambda_2 = 4 \lambda_3 = 9 \lambda_4$
7. As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom / ion [JEE Main 2015]
a. its kinetic energy increases but potential energy and total energy decrease
b. kinetic energy, potential energy and total energy decrease
c. kinetic energy decreases, potential energy increases but total energy remains same
d. kinetic energy and total energy decrease but potential energy increases
Answer: (a)
Solution:
For an electron in the hydrogen atom we have
$$\frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2}$$The centripetal force is provided by the electrostatic force of attraction.
$$\therefore mv^2 = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r}$$ $$\Rightarrow \text{KE} = \frac{1}{2}mv^2 = \frac{1}{8\pi\varepsilon_0} \frac{e^2}{r}$$The potential energy of the electron is given by
$$\text{PE} = -\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r}$$The total energy is given by
$$\text{TE} = \text{KE} + \text{PE} = \frac{1}{8\pi\varepsilon_0} \frac{e^2}{r} - \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r} = -\frac{1}{8\pi\varepsilon_0} \frac{e^2}{r}$$As the electron makes a transition from an excited state to the ground state of a hydrogen-like atom,
$r$ decreases, therefore kinetic energy increases as $\text{KE} = \frac{1}{8\pi\varepsilon_0} \frac{e^2}{r} \Rightarrow \text{KE} \propto \frac{1}{r}$
$r$ decreases, therefore potential energy decreases as $\text{PE} = -\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r} \Rightarrow \text{PE} \propto -\frac{1}{r}$
$r$ decreases, therefore total energy decreases as $\text{TE} = -\frac{1}{8\pi\varepsilon_0} \frac{e^2}{r} \Rightarrow \text{TE} \propto -\frac{1}{r}$
8. Highly excited states for hydrogen-like atoms (also called Rydberg states) with nuclear charge Ze are defined by their principal quantum number n, where n >> 1. Which of the following statement (s) is (are) true? [JEE Adv. 2016 P-I]
a. Relative change in the radii of two consecutive orbital’s does not depend on Z.
b. Relative change in the radii of two consecutive orbital’s varies as 1/n
c. Relative change in the energy of two consecutive orbital’s varies as $1/n^3$
d. Relative change in the angular moment a of two consecutive orbital’s varies as 1/n
Answer: (a, b, d)
Solution:
$$r_n \propto \frac{n^2}{Z}; \quad \frac{\Delta r}{r} = \frac{\Delta (n^2)}{n^2} \propto \frac{1}{n}$$ $$\Rightarrow \text{so } \frac{\Delta r}{r} \propto \frac{1}{n}$$ $$\frac{\Delta L}{L} = \frac{\frac{nh}{2\pi}}{n L} \propto \frac{1}{n}$$9. Some energy levels of a molecule are shown in figure. The ratio of the wavelength $r = \frac{\lambda_1}{\lambda_2}$ is given by: [JEE Main 2017]
a. $r = \frac{2}{3}$
b. $r = \frac{3}{4}$
c. $r = \frac{1}{3}$
d. $r = \frac{4}{3}$
Answer: (c)
Solution:
Using $\Delta E = \frac{hc}{\lambda}$
For $\lambda_1$:
$$-E - (-2E) = \frac{hc}{\lambda_1} \Rightarrow E = \frac{hc}{\lambda_1}$$For $\lambda_2$:
$$-\frac{4}{3}E - (-3E) = \frac{hc}{\lambda_2} \Rightarrow \frac{5}{3}E = \frac{hc}{\lambda_2}$$ $$\Rightarrow r = \frac{\lambda_1}{\lambda_2} = \frac{1}{3}$$10. If the series limit frequency of the Lyman series is vL, then the series limit frequency of the Pfund series is [JEE Main 2018]
a. $v_L / 25$
b. $25 v_L$
c. $16 v_L$
d. $v_L / 16$
Answer: (a)
Solution:
$$h v_L = E_1 \left[\frac{1}{1^2} - \frac{1}{\infty^2}\right] = E_1$$ $$h v_P = E_1 \left[\frac{1}{5^2} - \frac{1}{\infty^2}\right] = \frac{E_1}{25} \Rightarrow v_P = \frac{v_L}{25}$$11. A particle is moving in a circular path of radius a under the action of an attractive potential $U = -\frac{k}{2r^2}$. Its total energy is [JEE Main 2018]
a. $-\frac{3}{2}\frac{k}{a^2}$
b. $-\frac{k}{a^4}$
c. $-\frac{k}{a^2}$
d. Zero
Answer: (d)
Solution:
$$F = -\frac{dU}{dr}$$ $$U = \left[-\frac{k}{2r^2}\right]$$ $$\frac{mv^2}{r} = \frac{k}{r^3}$$[This force provides necessary centripetal force]
$$\Rightarrow mv^2 = \frac{k}{r^2}$$ $$\Rightarrow \text{KE} = \frac{k}{2r^2}$$ $$\Rightarrow \text{PE} = -\frac{k}{2r^2}$$ $$\text{Total energy} = \text{Zero}$$12. In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be: [JEE Main 2019]
a. $\frac{27}{20} \lambda$
b. $\frac{16}{25} \lambda$
c. $\frac{20}{27} \lambda$
d. $\frac{25}{16} \lambda$
Answer: (c)
Solution:
For M $\rightarrow$ L shell:
$$\frac{1}{\lambda} = K \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = K \times \frac{5}{36}$$For N $\rightarrow$ L shell:
$$\frac{1}{\lambda'} = K \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = K \times \frac{3}{16}$$ $$\lambda' = \frac{20}{27}\lambda$$13. A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980Ã…. The radius of the atom in the excited state, it terms of Bohr radius $a_0$, will be: (hc = 12500 eV – Ã…) [JEE Main 2019]
a. $9a_0$
b. $25a_0$
c. $4a_0$
d. $16a_0$
Answer: (d)
Solution:
$$\text{Energy of photon} = \frac{12500}{980} = 12.75\text{ eV}$$ $$\text{Electron will excite to } n = 4$$ $$\text{Since } r \propto n^2$$ $$\therefore \text{Radius of atom will be } 16a_0$$14. A particle of mass m moves in a circular orbit in a central potential field $U(r) = \frac{1}{2} K r^2$. If Bohr's quantisation conditions are applied, radii of possible orbital’s and energy levels vary with quantum number n as: [JEE Main 2019]
a. $r_n \propto n^2, E_n \propto n^2$
b. $r_n \propto n, E_n \propto n$
c. $r_n \propto n, E_n \propto n^2$
d. $r_n \propto n^{1/2}, E_n \propto n$
Answer: (d)
Solution:
$$F = \frac{dV}{dr} = \frac{mv^2}{r} = kr$$ $$mvr = \frac{nh}{2\pi}$$ $$r_n \propto n^{1/2}$$ $$E_n = \text{PE} + \text{KE} = \frac{1}{2}Kr^2 + \frac{1}{2}mv^2 \propto r^2 \propto n$$15. A free hydrogen atom after absorbing a photon of wavelength $\lambda_a$ gets excited from state n = 1 to n = 4 immediately after electron jumps to n = m state by emitting a photon of wavelength $\lambda_e$. Let change in momentum of atom due to the absorption and the emission are $\Delta P_a$ and $\Delta P_e$ respectively. If $\lambda_a / \lambda_e = 1/5$. Which of the following is correct? [JEE Adv. 2019 P-II]
a. m = 2
b. $\Delta P_a / \Delta P_e = 1/2$
c. $\lambda_e = 418\text{ nm}$
d. Ratio of K.E. of electron in the state n = m to n = 1 is 1/4
Answer: (a, d)
Solution:
$$\frac{\lambda_a}{\lambda_e} = \frac{E_4 - E_m}{E_4 - E_1} = \frac{\frac{1}{m^2} - \frac{1}{16}}{1 - \frac{1}{16}} = \frac{1}{5}$$On solving $m = 2$
$$\lambda_e = \frac{12400}{13.6 \times \left(\frac{1}{4} - \frac{1}{16}\right)} = 4340\text{ Ã…} = 434\text{ nm}$$ $$\frac{K_m}{K_1} = \frac{1/m^2}{1/1^2} = \frac{1}{4} \text{ as kinetic energy is proportional to } \frac{1}{n^2}$$Matrix Match/Column Match
16. Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II and indicate your answer by darkening appropriate bubbles in the $4 \times 4$ matrix given in the ORS. [JEE 2007 P-I]
Column I
(A) Transition between two atomic energy levels
(B) Electron emission from a material
(C) Moseley’s law
(D) Change of photon energy into kinetic energy of electrons
Column II
1. Characteristic X-rays
2. Photoelectric effect
3. Hydrogen spectrum
4. $\beta$-decay
a. A$\rightarrow$1,3; B$\rightarrow$2,4; C$\rightarrow$1; D$\rightarrow$2
b. A$\rightarrow$2,4; B$\rightarrow$1,2; C$\rightarrow$2; D$\rightarrow$3
c. A$\rightarrow$3,4; B$\rightarrow$3,4; C$\rightarrow$3; D$\rightarrow$1
d. A$\rightarrow$2,3; B$\rightarrow$2,3; C$\rightarrow$4; D$\rightarrow$2
Answer: (a)
Solution:
A$\rightarrow$1,3; B$\rightarrow$2,4; C$\rightarrow$1; D$\rightarrow$2
17. Match Column I (Electromagnetic wave type) with Column II (Its association application) and select the correct option from the choices given below the lists: [JEE Main 2014]
Column I
(A) Infrared waves
(B) Radio waves
(C) X-rays
(D) Ultraviolet rays
Column II
1. To treat muscular strain
2. For broadcasting
3. To detect fracture of bones
4. Absorbed by the ozone layer of the atmosphere
a. A$\rightarrow$1; B$\rightarrow$2; C$\rightarrow$3; D$\rightarrow$4
b. A$\rightarrow$3; B$\rightarrow$4; C$\rightarrow$1; D$\rightarrow$2
c. A$\rightarrow$2; B$\rightarrow$3; C$\rightarrow$2; D$\rightarrow$3
d. A$\rightarrow$4; B$\rightarrow$1; C$\rightarrow$2; D$\rightarrow$3
Answer: (a)
Solution:
A$\rightarrow$1; B$\rightarrow$2; C$\rightarrow$3; D$\rightarrow$4
Statement or Assertion and Reason
18. Statement-I: If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X-rays do not change. And Statement-II: When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy. [JEE 2007 P-I]
a. Statement-I is True, Statement-II is True; Statement -II is a correct explanation for Statement-I.
b. Statement-I is True, Statement-II is True; Statement -II is not a correct explanation for Statement-I.
c. Statement -I is True, Statement-II is False.
d. Statement -I is False, Statement-II is True.
Answer: (b)
Paragraphs
Paragraph for Question Nos. 19 to 21
In a mixture of H-$\text{He}^+$ gas ($\text{He}^+$ is singly ionised He atom), H atoms and $\text{He}^+$ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to $\text{He}^+$ ions (by collisions). Assume that the Bohr model of atom is exactly valid. [JEE 2008 P-I]
19. The quantum number n of the state finally populated in $\text{He}^+$ ions is
a. 2
b. 3
c. 4
d. 5
Answer: (c)
20. The wavelength of light emitted in the visible region by $\text{He}^+$ ions after collisions with H atom is:
a. $6.5 \times 10^{-7}\text{ m}$
b. $5.6 \times 10^{-7}\text{ m}$
c. $4.8 \times 10^{-7}\text{ m}$
d. $4.0 \times 10^{-7}\text{ m}$
Answer: (c)
Solution:
$$E_4 - E_3 = \frac{hc}{\lambda} \quad [\text{in visible region}]$$ $$\lambda = 4.8 \times 10^{-7}\text{ m}$$21. The ratio of the kinetic energy of the $n = 2$ electron for the H atom to that of $\text{He}^+$ ion is:
a. $\frac{1}{4}$
b. $\frac{1}{2}$
c. 1
d. 2
Answer: (a)
Solution:
$$\text{KE} \propto \frac{Z^2}{n^2} \Rightarrow \frac{\text{KE}_{\text{H}}}{\text{KE}_{\text{He}^+}} = \left(\frac{Z_{\text{H}}}{Z_{\text{He}^+}}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$Paragraph for Question Nos. 22 to 24
The key feature of Bohr’s theory of spectrum of hydrogen atom is the quantisation of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantised rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr’s quantisation condition.
22. A diatomic molecule has moment of inertia I. By Bohr’s quantization condition its rotational energy in the nth level (n = 0 is not allowed) is: [JEE 2010 P-II]
a. $\left(\frac{1}{8\pi^2}\right) \frac{h^2}{nI}$
b. $\left(\frac{1}{8\pi^2}\right) \frac{h}{nI}$
c. $\left(\frac{1}{8\pi^2}\right) \frac{h^2 n}{I}$
d. $\left(\frac{1}{8\pi^2}\right) \frac{n^2 h^2}{I}$
Answer: (d)
Solution:
$$L = \frac{nh}{2\pi} \Rightarrow \text{K.E.} = \frac{L^2}{2I} = \frac{1}{2I} \left(\frac{nh}{2\pi}\right)^2 = \left(\frac{1}{8\pi^2}\right)\frac{n^2 h^2}{I}$$23. It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to $\frac{4}{\pi} \times 10^{11}\text{ Hz}$. Then, the moment of inertia of CO molecule bout its centre of mass is close to (Take $h = 2\pi \times 10^{-34}\text{ Js}$)
a. $2.76 \times 10^{-46}\text{ kg m}^2$
b. $1.87 \times 10^{-46}\text{ kg m}^2$
c. $4.67 \times 10^{-47}\text{ kg m}^2$
d. $1.17 \times 10^{-47}\text{ kg m}^2$
Answer: (c)
Solution:
$$E_1 - E_0 = h \nu = \frac{h^2}{8\pi^2 I} (1^2 - 0)$$ $$I = 4.67 \times 10^{-47}\text{ kg m}^2$$24. In a CO molecule, the distance between C (mass = 12 a.m.u) and O (mass = 16 a.m.u.), where $1\text{ a.m.u.} = \frac{5}{3} \times 10^{-27}\text{ kg}$, is close to
a. $2.4 \times 10^{-10}\text{ m}$
b. $1.9 \times 10^{-10}\text{ m}$
c. $1.3 \times 10^{-10}\text{ m}$
d. $4.4 \times 10^{-10}\text{ m}$
Answer: (c)
Solution:
$$\mu = \frac{m_1 m_2}{m_1 + m_2}$$ $$I = \mu d^2$$ $$\therefore d = 1.3 \times 10^{-10}\text{ m}$$Integer and Subjective
25. In hydrogen-like atom $n^{\text{th}}$ line of Lyman series has wavelength equal to the de-Broglie’s wavelength of electron in the level from which it originated. What is the value of n? [JEE 2006]
Solution:
$$\lambda_{\text{de-Broglie}} = \frac{h}{p} = \frac{h}{m v_{n+1}}$$ $$\frac{1}{\lambda} = R Z^2 \left(\frac{1}{1^2} - \frac{1}{(n+1)^2}\right)$$Equating $\lambda_{\text{photon}} = \lambda_{\text{de-Broglie}}$, we get $n = 2$.
26. Consider a hydrogen atom with its electron in the orbital. An electromagnetic radiation of wavelength 90 nm is used to ionise the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of n is [JEE Adv. 2015 P-I]
Solution:
$$E_{\text{photon}} = \frac{hc}{\lambda} = \frac{1240}{90} \approx 13.77\text{ eV}$$ $$\text{KE} = E_{\text{photon}} - E_{\text{binding}} = 10.4\text{ eV}$$ $$\Rightarrow E_{\text{binding}} = 13.77 - 10.4 = 3.37\text{ eV}$$ $$\Rightarrow \frac{13.6}{n^2} \approx 3.4 \Rightarrow n = 2$$27. A hydrogen atom in its ground state is irradiated by light of wavelength 970 Ã…. Taking $hc/e = 1.237 \times 10^{-6}\text{ eV m}$ and the ground state energy of hydrogen atom as –13.6 eV, the number of lines present in the emission spectrum is: [JEE Adv. 2016 P-I]
Solution:
$$E_{\text{photon}} = \frac{12370}{970} \approx 12.75\text{ eV}$$ $$E_n - E_1 = -13.6 \left(\frac{1}{n^2} - 1\right) = 12.75$$ $$\Rightarrow \frac{1}{n^2} = 1 - \frac{12.75}{13.6} = \frac{0.85}{13.6} = \frac{1}{16} \Rightarrow n = 4$$ $$\text{No. of lines} = {}^4\text{C}_2 = \frac{4 \times 3}{2} = 6$$28. An electron in hydrogen atom undergoes a transition from an orbit with quantum number $n_i$ to another with quantum number $n_f$. $V_i$ and $V_f$ are respectively the initial and final potential energies of the electron. If $\frac{V_i}{V_f} = 6.25$, then the smallest possible $n_f$ is [JEE Adv. 2017 P-I]
Solution:
Potential energy of electron in $n^{\text{th}}$ orbit of hydrogen atom,
$$U_n = -\frac{27.2}{n^2}\text{ eV}$$ $$\frac{U_i}{U_f} = \frac{n_f^2}{n_i^2} = 6.25$$ $$\Rightarrow \frac{n_f}{n_i} = 2.5 = \frac{5}{2}$$ $$\text{So, } (n_f)_{\min} = 5$$29. Consider a hydrogen-like ionised atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionisation energy of the hydrogen atom is 13.6 eV. The value of Z is ______. [JEE Adv. 2018 P-II]
Answer: 3
Solution:
$$13.6 Z^2 \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 74.8 + 13.6 Z^2 \left(\frac{1}{2^2} - \frac{1}{3^2}\right)$$ $$\Rightarrow 13.6 Z^2 \left(\frac{3}{4}\right) - 13.6 Z^2 \left(\frac{5}{36}\right) = 74.8$$ $$\Rightarrow 13.6 Z^2 \left(\frac{22}{36}\right) = 74.8$$ $$\Rightarrow Z^2 = 9 \Rightarrow Z = 3$$
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