#

Multiple Choice Questions

1. Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic molecule $\text{B}_2$ is: [JEE 2010 P-II]

(a) 1 and diamagnetic

(b) 0 and diamagnetic

(d) 0 and paramagnetic

Answer: (a)

Explanation: $\text{B}_2 (10) = \sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \pi 2p_x^2$ (due to violation of Hund's rule).

$$\text{Bond order} = \frac{6 - 4}{2} = 1$$

Nature is diamagnetic as there are no unpaired electrons.

2. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [$a_0$ is Bohr radius]. [JEE 2012 P-I]

(a) $\frac{h^2}{4 \pi^2 m a_0^2}$

(b) $\frac{h^2}{16 \pi^2 m a_0^2}$

(d) $\frac{h^2}{64 \pi^2 m a_0^2}$

Answer: (c)

Explanation: $mvr = \frac{nh}{2\pi}$ and $\frac{mv^2}{r} = \frac{e^2}{r^2} \implies mv^2 r = e^2$.

$$\text{K.E.} = \frac{1}{2}mv^2 = \frac{me^4 \cdot 4\pi^2}{2 n^2 h^2}$$

Since $a_0 = \frac{h^2}{4\pi^2 m e^2} \implies e^2 = \frac{h^2}{4\pi^2 m a_0}$.

For $n = 2$:

$$\text{K.E.} = \frac{1}{2} \times \frac{h^2}{8 \pi^2 m a_0^2 \cdot 4} = \frac{h^2}{32 \pi^2 m a_0^2}$$

3. Which of the following represents the correct order of increasing first ionisation enthalpy for Ca, Ba, S, Se and Ar? [JEE Main 2013]

(a) $\text{Ca} < \text{S} < \text{Ba} < \text{Se} < \text{Ar}$

(b) $\text{S} < \text{Se} < \text{Ca} < \text{Ba} < \text{Ar}$

(d) $\text{Ca} < \text{Ba} < \text{S} < \text{Se} < \text{Ar}$

Answer: (c)

Explanation: Ionisation enthalpy increases across a period and decreases down a group. $\text{Ba}$ has lowest first ionisation enthalpy, while noble gas $\text{Ar}$ has highest. The correct increasing order is $\text{Ba} < \text{Ca} < \text{Se} < \text{S} < \text{Ar}$.

4. Energy of an electron is given by $E = -2.178 \times 10^{-18} \left(\frac{Z^2}{n^2}\right)\text{ J}$. Wavelength of light required to excite an electron in a hydrogen atom from level $n=1$ to $n=2$ will be ($h=6.62 \times 10^{-34}\text{ Js}$ and $c=3.0 \times 10^8\text{ ms}^{-1}$): [JEE Main 2013]

(a) $1.214 \times 10^{-7}\text{ m}$

(b) $2.816 \times 10^{-7}\text{ m}$

(d) $8.500 \times 10^{-7}\text{ m}$

Answer: (a)

Explanation: $\frac{1}{\lambda} = R_H Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$. For $Z = 1$, $n_1 = 1$, $n_2 = 2$:

$$\lambda = 1.214 \times 10^{-7}\text{ m}$$

5. The first ionisation potential of Na is 5.1 eV. The value of electron gain enthalpy of $\text{Na}^+$ will be: [JEE Main 2013]

(a) $-2.55\text{ eV}$

(b) $-5.1\text{ eV}$

(d) $+2.55\text{ eV}$

Answer: (b)

Explanation: $\text{E.A. of Na}^+ = -\text{I.P. of Na} = -5.1\text{ eV}$.

6. Which of the following is the energy of a possible excited state of hydrogen? [JEE Main 2015]

(a) $+13.6\text{ eV}$

(b) $-6.8\text{ eV}$

(d) $+6.8\text{ eV}$

Answer: (c)

Explanation: Energy in $1^{\text{st}}$ excited state ($n=2$) is:

$$E_2 = -\frac{13.6}{2^2} = -3.4\text{ eV}$$

7. The ionic radii (in Å) of $\text{N}^{3-}$, $\text{O}^{2-}$ and $\text{F}^-$ are respectively: [JEE Main 2015]

(a) 1.36, 1.40 and 1.71

(b) 1.36, 1.71 and 1.40

(d) 1.71, 1.36 and 1.40

Answer: (c)

Explanation: For isoelectronic species, higher negative charge gives larger radius. Ionic radii order: $\text{N}^{3-} > \text{O}^{2-} > \text{F}^-$. Thus, the radii are 1.71, 1.40 and 1.36 Å respectively.

8. A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference $V\text{ esu}$. If $e$ and $m$ are charge and mass of an electron, respectively, then the value of $\frac{h}{\lambda}$ (where $\lambda$ is wavelength associated with electron wave) is given by: [JEE Main 2016]

(a) $meV$

(b) $2meV$

(d) $\sqrt{2meV}$

Answer: (d)

Explanation: $\text{K.E.} = eV$. de Broglie wavelength $\lambda = \frac{h}{\sqrt{2m \cdot \text{K.E.}}} = \frac{h}{\sqrt{2meV}} \implies \frac{h}{\lambda} = \sqrt{2meV}$.

9. Which of the following atoms has the highest first ionisation energy? [JEE Main 2016]

(a) Rb

(b) Na

(d) Sc

Answer: (d)

Explanation: First ionisation potential order: $\text{Sc} > \text{Na} > \text{K} > \text{Rb}$.

10. $P$ is the probability of finding the 1s electron of hydrogen atom in a spherical shell of infinitesimal thickness, $dr$, at a distance $r$ from the nucleus. The volume of this shell is $4\pi r^2 dr$. The qualitative sketch of the dependence of $P$ on $r$ is: [JEE Adv. 2016 P-I]

(a) A curve starting from positive $P$ value at $r=0$ and decaying exponentially.

(b) A curve starting from zero, rising to a peak, and then exponentially decaying.

(d) A curve starting from zero and reaching a maximum value at $r=0$.

Answer: (d)

Explanation: For 1s electron in H-atom, plot of radial probability function $P = 4\pi r^2 R^2$ vs $r$ starts from zero at $r=0$, reaches a maximum at $r = a_0$, and then decreases to zero.

11. The radius of the second Bohr orbit for hydrogen atom is: (Planck's const. $h = 6.6262 \times 10^{-34}\text{ Js}$; Mass of electron = $9.1091 \times 10^{-31}\text{ kg}$; charge of electron $e = 1.60210 \times 10^{-19}\text{ C}$; permittivity of vacuum $\epsilon_0 = 8.854185 \times 10^{-12}\text{ kg}^{-1}\text{ m}^{-3}\text{ A}^2\text{ s}^4$) [JEE Main 2017]

(a) 2.12 Å

(b) 1.65 Å

(d) 0.529 Å

Answer: (a)

Explanation: Radius of $n^{\text{th}}$ Bohr orbit in H-atom $= 0.53 n^2\text{ Å}$.

Radius of $2^{\text{nd}}$ Bohr orbit $= 0.53 \times 2^2 = 2.12\text{ Å}$.

12. According to molecular orbital, which of the following will not be a viable molecule? [JEE Main 2018]

(a) $\text{H}_2^{2-}$

(b) $\text{He}_2^{2+}$

(d) $\text{H}_2^-$

Answer: (a)

Explanation: Electronic configuration and bond order:

$\text{He}_2^+ (\sigma 1s^2 \sigma^* 1s^1)$, $\text{B.O.} = \frac{2 - 1}{2} = 0.5$

$\text{H}_2^- (\sigma 1s^2 \sigma^* 1s^1)$, $\text{B.O.} = \frac{2 - 1}{2} = 0.5$

$\text{H}_2^{2-} (\sigma 1s^2 \sigma^* 1s^2)$, $\text{B.O.} = \frac{2 - 2}{2} = 0$

$\text{He}_2^{2+} (\sigma 1s^2)$, $\text{B.O.} = \frac{2 - 0}{2} = 1$

A molecule having zero bond order will not be a viable molecule.

13. The correct statement(s) regarding the binary transition metal carbonyl compounds is(are) (Atomic numbers: $\text{Fe} = 26, \text{Ni} = 28$) [JEE Adv. 2018 P-I]

(a) Total number of valence shell electrons at metal centre in $\text{Fe(CO)}_5$ or $\text{Ni(CO)}_4$ is 16

(b) These are predominantly low spin in nature

(d) The carbonyl $\text{C-O}$ bond weakens when the oxidation state of the metal is increased

Answer: (b), (c)

Explanation: Valence shell electrons in $\text{Fe(CO)}_5 = 8 + 2 \times 5 = 18$ electrons. In $\text{Ni(CO)}_4 = 10 + 2 \times 4 = 18$ electrons. Both complexes are formed in low spin condition due to strong field ligand $\text{CO}$. Metal-carbon bond strengthens when complex is formed in lower oxidation number of metal due to extent of back-bonding ($\pi$-back donation).

14. The quantum number of four electrons are given below:

I. $n = 4, l = 2, m_l = -2, m_s = -\frac{1}{2}$

II. $n = 3, l = 2, m_l = 1, m_s = +\frac{1}{2}$

III. $n = 4, l = 1, m_l = 0, m_s = +\frac{1}{2}$

IV. $n = 3, l = 1, m_l = 1, m_s = -\frac{1}{2}$

The correct order of their increasing energies will be: [JEE Main 2019]

(a) IV < III < II < I

(b) IV < II < III < I

(d) I < III < II < IV

Answer: (b)

Explanation: According to $(n + l)$ rule: IV ($3p$, $n+l=4$) < II ($3d$, $n+l=5$) < III ($4p$, $n+l=5$) < I ($4d$, $n+l=6$).

15. If $p$ is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength $\lambda$, then for $1.5 p$ momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function) [JEE Main 2019]

(a) $\frac{1}{2}\lambda$

(b) $\frac{3}{4}\lambda$

(d) $\frac{4}{9}\lambda$

Answer: (d)

Explanation: $h\nu = \text{K.E.} + \phi \approx \text{K.E.}$, so $\frac{hc}{\lambda} \approx \text{K.E.} = \frac{p^2}{2m}$.

$$\frac{p^2}{2m} = \frac{hc}{\lambda} \quad \text{--- (i)}$$

$$\frac{(1.5 p)^2}{2m} = \frac{hc}{\lambda'} \quad \text{--- (ii)}$$

Dividing (i) by (ii):

$$\frac{1}{(1.5)^2} = \frac{\lambda'}{\lambda} \implies \lambda' = \frac{4}{9}\lambda$$

16. For emission line of atomic hydrogen from $n_1 = 8$ to $n_f$, the plot of wave number ($\bar{\nu}$) against $\frac{1}{n^2}$ will be ($R_H$ is Rydberg constant): [JEE Main 2019]

(a) Linear with slope $-R_H$

(b) Linear with intercept $-R_H$

(d) Linear with slope $R_H$

Answer: (d)

Explanation: $\bar{\nu} = R_H Z^2 \left(\frac{1}{n_f^2} - \frac{1}{8^2}\right) = R_H \cdot \left(\frac{1}{n_f^2}\right) - \frac{R_H}{64}$.

Comparing with $y = mx + c$, slope $m = R_H$ (Linear with slope $R_H$).

17. The highest value of the calculated spin only magnetic moment (in BM) among all the transition metal complexes is: [JEE Main 2019]

(a) 5.92

(b) 3.87

(d) 4.90

Answer: (a)

Explanation: $\mu = \sqrt{n(n+2)}\text{ B.M.}$. Maximum number of unpaired electrons in $d$-orbitals $= 5$ (e.g., $\text{Mn}^{2+}$ complex).

$$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\text{ B.M.}$$

18. Which of the following combination of statements is true regarding the interpretation of the atomic orbitals?

1. An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum.

2. For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number.

3. According to wave mechanics, the ground state angular momentum of $\text{H}$ equal to $\frac{h}{2\pi}$.

4. The plot of $\psi$ vs $r$ for various azimuthal quantum numbers, shown peak shifting towards higher $r$ value. [JEE Main 2019]

(a) 2, 3

(b) 1, 4

(d) 1, 3

Answer: (b)

Explanation: Statements 1 and 4 correctly describe radial probability distribution behavior with azimuthal quantum numbers.

19. The metal used for making X-ray tube window is: [JEE Main 2019]

(a) Mg

(b) Na

(d) Be

Answer: (d)

Explanation: $\text{Be}$ metal is used in X-ray tube windows because it is transparent to X-rays.

20. Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from metal surface? [JEE Main 2019]

(a) K.E. vs Frequency of light (Linear graph with positive slope and intercept on frequency axis)

(b) K.E. vs Intensity of light (Horizontal line, zero slope)

(d) Number of $\text{e}^-$s vs Frequency of light (Horizontal line after threshold frequency)

Answer: (c)

Explanation: K.E. of ejected photoelectron depends on frequency according to $E = h\nu - h\nu_0$. It does not start directly from origin since work function threshold must be overcome. Thus, graph (c) is incorrect.

21. The ground state energy of hydrogen atom is –13.6 eV. The energy of second excited state $\text{He}^+$ ion in eV is: [JEE Main 2019]

(a) –6.04

(b) –27.2

(d) –3.4

Answer: (a)

Explanation: $E = -13.6 \times \frac{Z^2}{n^2}\text{ eV}$. For second excited state of $\text{He}^+$ ($Z=2, n=3$):

$$E_3(\text{He}^+) = -13.6 \times \frac{2^2}{3^2} = -13.6 \times \frac{4}{9} = -6.04\text{ eV}$$

22. Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H-atom is suitable for this purpose? [$R_H = 1 \times 10^5\text{ cm}^{-1}, h = 6.6 \times 10^{-34}\text{ Js}, c = 3 \times 10^8\text{ ms}^{-1}$] [JEE Main 2019]

(a) Paschen, $5 \rightarrow 3$

(b) Paschen, $\infty \rightarrow 3$

(d) Balmer, $\infty \rightarrow 2$

Answer: (b)

Explanation: Paschen limit ($\infty \rightarrow 3$) gives wavelength $\lambda = \frac{1}{R_H \left(\frac{1}{3^2} - 0\right)} = \frac{9}{10^5\text{ cm}^{-1}} = 9 \times 10^{-5}\text{ cm} = 900\text{ nm}$.

23. The de Broglie wavelength ($\lambda$) associated with a photoelectron varies with the frequency ($\nu$) of the incident radiation as, [$\nu_0$ is threshold frequency]: [JEE Main 2019]

(a) $\lambda \propto \frac{1}{(\nu - \nu_0)^{3/2}}$

(b) $\lambda \propto \frac{1}{(\nu - \nu_0)^{1/2}}$

(d) $\lambda \propto \frac{1}{(\nu - \nu_0)}$

Answer: (b)

Explanation: $\text{K.E.} = h(\nu - \nu_0)$. de Broglie wavelength $\lambda = \frac{h}{\sqrt{2m \cdot \text{K.E.}}} = \frac{h}{\sqrt{2m h(\nu - \nu_0)}} \implies \lambda \propto \frac{1}{(\nu - \nu_0)^{1/2}}$.

24. What is the work function of the metal if the light of wavelength 4000 Å generates photoelectrons of velocity $6 \times 10^5\text{ ms}^{-1}$ from it? [JEE Main 2019] (Mass of electron $= 9 \times 10^{-31}\text{ kg}$, Velocity of light $= 3 \times 10^8\text{ ms}^{-1}$, Planck's constant $= 6.626 \times 10^{-34}\text{ Js}$, Charge of electron $= 1.6 \times 10^{-19}\text{ J eV}^{-1}$)

(a) 0.9 eV

(b) 4.0 eV

(d) 3.1 eV

Answer: (c)

Explanation: $h\nu = \phi + \frac{1}{2}mv^2 \implies \frac{hc}{\lambda} = \phi + \frac{1}{2}mv^2$.

$$\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4000 \times 10^{-10}} = \phi + \frac{1}{2} \times 9 \times 10^{-31} \times (6 \times 10^5)^2$$

$$\phi = 3.35 \times 10^{-19}\text{ J} \implies \phi = 2.1\text{ eV}$$

25. The upper stratosphere consisting of the ozone layer protects us from the sun's radiation that falls in the wavelength region of: [JEE Main 2019]

(a) 600-750 nm

(b) 0.8-1.5 nm

(d) 200-315 nm

Answer: (d)

Explanation: Ozone protects most of the medium frequencies ultraviolet light from 200 to 315 nm of wavelength.

26. If the de Broglie wavelength of the electron in $n^{\text{th}}$ Bohr orbit in a hydrogenic atom is equal to $1.5 \pi a_0$ ($a_0$ is Bohr radius), then the value of $n/Z$ is: [JEE Main 2019]

(a) 1.0

(b) 0.75

(d) 1.50

Answer: (b)

Explanation: According to de-Broglie's hypothesis: $2\pi r_n = n\lambda$.

$$2\pi a_0 \frac{n^2}{Z} = n (1.5 \pi a_0) \implies \frac{n}{Z} = \frac{1.5}{2} = 0.75$$

27. For $\text{He}^+$ the electron is in orbit with energy equal to $-3.4\text{ eV}$. The azimuthal quantum number for that orbit is 2 and magnetic quantum number is 0. Then which of the following is/are correct. [JEE Adv. 2019 P-II]

(a) The subshell is 4d.

(b) The number of angular nodes in it is 2.

(d) The nuclear charge experienced in $n = 4$ is $2e$ less than that in $n = 1$, where $e$ is electronic charge.

Answer: (a), (b)

Explanation: $E_n = -13.6 \times \frac{2^2}{n^2} = -3.4\text{ eV} \implies n^2 = 16 \implies n = 4$.

Quantum numbers: $n = 4, l = 2, m = 0 \implies$ subshell is 4d.

Angular nodes $= l = 2$. Radial nodes $= n - l - 1 = 4 - 2 - 1 = 1$.

Matrix Match / Column Match

28. According to Bohr’s theory, Total energy $E_n$, Kinetic energy $K_n$, Potential energy $V_n$, Radius of $n^{\text{th}}$ orbit $r_n$. Match the following: [JEE 2006]

Column I	Column II
(A) $V_n / K_n$	1. 0
(B) If radius of $n^{\text{th}}$ orbit $\propto E_n^x$, $x =$	2. –1
(C) Angular momentum in lowest orbital	3. –2
(D) $\frac{1}{r_n} \propto Z^y$, $y =$	4. 1

(a) A $\rightarrow$ 3; B $\rightarrow$ 2; C $\rightarrow$ 1; D $\rightarrow$ 4

(b) A $\rightarrow$ 1; B $\rightarrow$ 2; C $\rightarrow$ 3; D $\rightarrow$ 4

(d) A $\rightarrow$ 3; B $\rightarrow$ 4; C $\rightarrow$ 2; D $\rightarrow$ 1

Answer: (a)

Explanation: $V_n = -2 K_n \implies V_n / K_n = -2$. $r_n \propto n^2$ and $E_n \propto 1/n^2 \implies r_n \propto E_n^{-1} \implies x = -1$. Angular momentum of lowest orbital ($l=0$) is $0$. $r_n \propto 1/Z \implies \frac{1}{r_n} \propto Z^1 \implies y = 1$.

29. Match the entries in Column I with the correctly related quantum number(s) in Column II. [JEE 2008 P-II]

Column I	Column II
(A) Orbital angular momentum of the electron in a hydrogen-like atomic orbital	1. Principal quantum number
(B) A hydrogen-like one-electron wave function obeying Pauli principle	2. Azimuthal quantum number
(C) Shape, size and orientation of hydrogen-like atomic orbitals	3. Magnetic quantum number
(D) Probability density of electron at the nucleus in hydrogen-like atom	4. Electron spin quantum number

(a) A $\rightarrow$ 1; B $\rightarrow$ 1, 2, 3; C $\rightarrow$ 1, 4; D $\rightarrow$ 1, 2

(b) A $\rightarrow$ 2; B $\rightarrow$ 4; C $\rightarrow$ 1, 2, 3; D $\rightarrow$ 1, 2, 3

(d) A $\rightarrow$ 1, 3; B $\rightarrow$ 1, 2, 3, 4; C $\rightarrow$ 2; D $\rightarrow$ 1, 2, 3

Answer: (b)

Explanation: Orbital angular momentum depends on $l$ (2). Pauli principle specifies full state including spin $m_s$ (4). Shape ($l$), size ($n$), and orientation ($m_l$) depend on $n, l, m_l$ (1, 2, 3). Probability density at nucleus depends on $n, l, m_l$ (1, 2, 3).

Match the Columns for Question Nos. 30 to 32

The information is given in the three columns of the following table. [JEE Adv. 2017 P-I]

Column I	Column II	Column III
(I) 1s orbital	(i) $\psi_{n,l,m_l} \propto \left(\frac{Z}{a_0}\right)^{3/2} e^{-\frac{Zr}{a_0}}$	(A) Probability density $\propto \frac{1}{a_0^3}$
(II) 2s orbital	(ii) One radial node	(B) Probability density is maximum at nucleus
(III) $2p_z$ orbital	(iii) $\psi_{n,l,m_l} \propto \left(\frac{Z}{a_0}\right)^{5/2} r e^{-\frac{Zr}{2a_0}} \cos\theta$	(C) Energy needed to excite electron from $n=2$ state to $n=4$ state is $\frac{27}{32}$ times energy to excite from $n=2$ to $n=6$
(IV) $3d_{z^2}$ orbital	(iv) $xy$-plane is a nodal plane	(D) Energy needed to excite electron from $n=2$ state to $n=4$ state is $\frac{27}{32}$ times energy to excite from $n=2$ to $n=6$

30. For $\text{He}^+$ ion, the only incorrect combination is: [JEE Adv. 2017 P-I]

(a) (I) (i) (C)

(b) (II) (ii) (B)

(d) (I) (iii) (C)

Answer: (d)

Explanation: 1s orbital is non-directional, so $\psi$ will not depend upon $\cos\theta$. Hence (I) with (iii) is incorrect.

31. For the given orbital in Column I, the only correct combination for any hydrogen-like species is: [JEE Adv. 2017 P-I]

(a) (I) (ii) (D)

(b) (IV) (iv) (C)

(d) (II) (ii) (A)

Answer: (d)

Explanation: 2s orbital has one radial node ($n-l-1 = 2-0-1 = 1$) and probability density at nucleus $\propto a_0^{-3}$.

32. For hydrogen atom, the only correct combination is: [JEE Adv. 2017 P-I]

(a) (II) (i) (A)

(b) (I) (iv) (B)

(d) (I) (i) (D)

Answer: (d)

Explanation: For 1s orbital, wavefunction is $\psi_{1s} \propto \left(\frac{Z}{a_0}\right)^{3/2} e^{-\frac{Zr}{a_0}}$. $\Delta E(2 \rightarrow 4) = E_1 \left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3}{16} E_1$, $\Delta E(2 \rightarrow 6) = E_1 \left(\frac{1}{4} - \frac{1}{36}\right) = \frac{2}{9} E_1$. Ratio $= \frac{3/16}{2/9} = \frac{27}{32}$.

For Question Nos. 33 to 34

Match the Columns: [JEE Adv. 2019 P-II]

Column I	Column II
(A) Radius	1. $\propto n^{-1}$
(B) Angular momentum	2. $\propto n^{-2}$
(C) Kinetic energy	3. $\propto n^0$
(D) Potential energy	4. $\propto n^1$
	5. $\propto n^2$

33. Which of the following is correct?

(a) A $\rightarrow$ 1

(b) A $\rightarrow$ 2

(d) A $\rightarrow$ 3

Answer: (c)

Explanation: Radius $r_n \propto n^2 \implies \text{A} \rightarrow 5$.

34. Which of the following is correct?

(a) D $\rightarrow$ 4

(b) C $\rightarrow$ 1

(d) D $\rightarrow$ 3

Answer: (c)

Explanation: Kinetic energy $K.E. \propto \frac{1}{n^2} = n^{-2} \implies \text{C} \rightarrow 2$.

Paragraph

Paragraph for Question Nos. 35 to 37

The hydrogen-like species $S_1$ is in a spherically symmetric state with one radial node. Upon absorbing light the ion undergoes transition to a state $S_2$. The state $S_2$ has one radial node and its energy is equal to the ground state energy of the hydrogen atom. [JEE 2010 P-II]

35. The state $S_1$ is:

(a) 1s

(b) 2s

(d) 3s

Answer: (b)

Explanation: For $S_1$, state is spherically symmetric ($l=0$, s-orbital) and has one radial node ($n-l-1 = 1 \implies n - 1 = 1 \implies n=2$). So $S_1$ is 2s.

36. Energy of the state $S_1$ in units of the hydrogen atom ground state energy is:

(a) 0.75

(b) 1.50

(d) 4.50

Answer: (c)

Explanation: For $S_2$, radial node $= 1 \implies n_2 - l_2 - 1 = 1$. Since energy of $S_2$ equals ground state energy of H-atom ($-13.6\text{ eV}$):

$$-13.6 \times \frac{Z^2}{n_2^2} = -13.6 \implies Z = n_2$$

For transition from 2s to $S_2$ with absorption of light, $l_2 = 1 \implies n_2 = 3 \implies Z = 3$.

Energy of $S_1$ (2s): $E(S_1) = E_H \times \frac{Z^2}{2^2} = E_H \times \frac{9}{4} = 2.25 E_H$.

37. The orbital angular momentum quantum number of the state $S_2$ is:

(a) 0

(b) 1

(d) 3

Answer: (b)

Explanation: Selection rule for dipole absorption transition is $\Delta l = \pm 1$. Since $S_1$ is $2s$ ($l=0$), the excited state $S_2$ must have $l=1$ (p-orbital).

Integer and Subjective

38. The maximum number of electrons that can have principal quantum number $n = 3$, and spin quantum number $m_s = -\frac{1}{2}$, is: [JEE 2011 P-I]

Answer: 9

Explanation: Total orbitals for $n = 3$ is $n^2 = 3^2 = 9$. Each orbital can have 1 electron with spin $m_s = -\frac{1}{2}$. Thus, maximum electrons $= 9$.

39. The work function of some metals is listed below. The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal is: [JEE 2011 P-I]

Metal	Li	Na	K	Mg	Cu	Ag	Fe	Pt	W
$\phi$ (eV)	2.4	2.3	2.2	3.7	4.8	4.3	4.7	6.3	4.75

Answer: 4

Explanation: Energy of incident photon $E = \frac{hc}{\lambda} = \frac{1240}{300\text{ nm}} = 4.13\text{ eV}$. Photoelectric effect occurs when $E \ge \phi$. Metals with $\phi \le 4.13\text{ eV}$ are Li (2.4), Na (2.3), K (2.2), and Mg (3.7). Total number of metals $= 4$.

40. The atomic masses of He and Ne are 4 and 20 a.m.u., respectively. The value of the de Broglie wavelength of He gas at $-73^\circ\text{C}$ is "M" times that of the de Broglie wavelength of Ne at $727^\circ\text{C}$. M is: [JEE Adv. 2013 P-I]

Answer: 5

Explanation: $\lambda = \frac{h}{mv} = \frac{h}{\sqrt{3mKT}} \implies \lambda \propto \frac{1}{\sqrt{mT}}$.

$$\frac{\lambda_{\text{He}}}{\lambda_{\text{Ne}}} = \sqrt{\frac{m_{\text{Ne}} T_{\text{Ne}}}{m_{\text{He}} T_{\text{He}}}} = \sqrt{\frac{20 \times (727 + 273)}{4 \times (-73 + 273)}} = \sqrt{\frac{20 \times 1000}{4 \times 200}} = \sqrt{25} = 5$$

41. In an atom, the total number of electrons having quantum number $n = 4, |m_l| = 1$ and $m_s = -\frac{1}{2}$ is: [JEE Adv. 2014 P-I]

Answer: 6

Explanation: For $n = 4$:

$l = 1 (4p): m_l = -1, +1 \implies 2\text{ orbitals}$

$l = 2 (4d): m_l = -1, +1 \implies 2\text{ orbitals}$

$l = 3 (4f): m_l = -1, +1 \implies 2\text{ orbitals}$

Total orbitals with $|m_l| = 1$ is $2 + 2 + 2 = 6$. Each orbital has 1 electron with $m_s = -\frac{1}{2}$. Total electrons $= 6$.

42. Not considering the electronic spin, the degeneracy of the second excited state ($n = 3$) of H atom is 9, while the degeneracy of the second excited state of $\text{H}^-$ is: [JEE Adv. 2015 P-I]

Answer: 3

Explanation: $\text{H}^-$ is a two-electron system. Ground state: $1s^2$. First excited state: $1s^1 2s^1, 1s^1 2p^1$. Second excited state: $1s^1 3s^1, 1s^1 3p^3, 1s^1 3d^5$? Wait, for multielectron species, degeneracy of $2p$ configuration in excited state $1s^1 2p^1$ is 3 ($2p_x, 2p_y, 2p_z$). Thus, degeneracy is 3.

43. (a) Calculate velocity of electron in first Bohr orbit of hydrogen atom (Given $r_0 = a_0 = 0.529\text{ Å}$).

(b) Find de-Broglie wavelength of the electron in first Bohr orbit.

Answer: (a) $2.18 \times 10^6\text{ m/s}$, (b) $3.3\text{ Å}$, (c) $\sqrt{2} \frac{h}{2\pi}$

Explanation:

(a) $mvr = \frac{nh}{2\pi} \implies v = \frac{nh}{2\pi m r} = \frac{1 \times 6.63 \times 10^{-34}}{2\pi \times 9.1 \times 10^{-31} \times 0.529 \times 10^{-10}} = 2.18 \times 10^6\text{ m/sec}$.

(b) $\lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 2.18 \times 10^6} = 0.33 \times 10^{-9}\text{ m} = 3.3\text{ Å}$.

(c) Orbital angular momentum $= \sqrt{l(l+1)}\frac{h}{2\pi}$. For 2p orbital, $l = 1 \implies \sqrt{1(1+1)}\frac{h}{2\pi} = \sqrt{2}\frac{h}{2\pi}$.