JEE Physics Atoms MCQs Set 11

Answer: (b) \( \frac{ke^2}{2r} \)

Explanation:
1. The Coulomb force supplies the centripetal force: \( \frac{mv^2}{r}=\frac{ke^2}{r^2} \).
2. So \( K=\frac12mv^2=\frac{ke^2}{2r} \).

Teacher's Note:
1. Total energy is -ke^2/2r and potential energy is -ke^2/r.

Answer: (d) \( \frac25v \)

Explanation:
1. \( v_n\propto\frac Zn \) for a given nucleus.
2. So \( \frac{v_5}{v_2}=\frac25 \).

Teacher's Note:
1. Speed falls as 1/n.

Answer: (a) I and III only

Explanation:
1. \( v\propto\frac1n \) so (I) is true.
2. \( r\propto n^2 \), not n, so (II) is false.
3. \( f=\frac{v}{2\pi r}\propto\frac1{n^3} \) so (III) is true; the force \( \propto\frac1{r^2} \) decreases outward so (IV) is false.

Teacher's Note:
1. Remember v ~ 1/n, r ~ n^2, f ~ 1/n^3.

Answer: (d) di-ionised lithium

Explanation:
1. The energy released is proportional to \( Z^2 \), and shorter wavelength means larger energy.
2. Li\( ^{2+} \) has Z = 3, the largest of the one-electron species given (H: 1, D: 1, He\( ^+ \): 2).

Teacher's Note:
1. Hydrogen-like energy levels scale as Z^2.

Answer: (b) Balmer series

Explanation:
1. Lyman lines are in the ultraviolet.
2. Balmer lines (ending at n = 2) are visible; Paschen and Brackett are infrared.

Teacher's Note:
1. Visible lines: red 656 nm, blue-green 486 nm, and so on.

Answer: (a) potential energy increases and kinetic energy decreases

Explanation:
1. \( K=\frac{13.6}{n^2} \) decreases with n.
2. Total energy rises (less negative) and \( U=-2K \), so potential energy increases.

Teacher's Note:
1. K = -E and U = 2E for a Bohr orbit.

Answer: (a) \( -1 \)

Explanation:
1. For Bohr orbits, \( K=-E \) (the total energy is negative).
2. So \( K/E=-1 \).

Teacher's Note:
1. U = 2E, K = -E.

Answer: (c) 5 : 27

Explanation:
1. \( \frac1{\lambda_L}\propto1-\frac14=\frac34 \) and \( \frac1{\lambda_B}\propto\frac14-\frac19=\frac5{36} \).
2. \( \frac{\lambda_L}{\lambda_B}=\frac{5/36}{3/4}=\frac5{27} \).

Teacher's Note:
1. Larger energy gap means smaller wavelength.

Answer: (d) 0.66 eV

Explanation:
1. \( E=13.6\left(rac19-rac1{16} ight)=13.6 imesrac7{144} \).
2. This is 0.66 eV.

Teacher's Note:
1. Use E = 13.6 (1/n1^2 - 1/n2^2) eV.

Answer: (b) \( \frac{20}{27}\lambda_0 \)

Explanation:
1. \( \frac1{\lambda_0}\propto\frac14-\frac19=\frac5{36} \) and \( \frac1{\lambda}\propto\frac14-\frac1{16}=\frac3{16} \).
2. \( \lambda=\lambda_0\times\frac{5/36}{3/16}=\frac{20}{27}\lambda_0 \).

Teacher's Note:
1. A larger gap gives a shorter wavelength.

Answer: (b) 30.6 eV

Explanation:
1. For Li\( ^{2+} \), \( Z=3 \) and the first excited state is n = 2.
2. Energy required \( =\frac{13.6\times9}{4}=30.6 \) eV.

Teacher's Note:
1. Use 13.6 Z^2 / n^2.

Answer: (b) -27.2 eV

Explanation:
1. \( U=2E \) for a Bohr orbit.
2. So \( U=2 imes(-13.6)=-27.2 \) eV.

Teacher's Note:
1. Kinetic +13.6 eV, potential -27.2 eV, total -13.6 eV.

Answer: (a) \( 2\pi r \)

Explanation:
1. Bohr's condition is equivalent to \( 2\pi r=n\lambda \).
2. For n = 1, \( \lambda=2\pi r \).

Teacher's Note:
1. The orbit holds a whole number of wavelengths.

Answer: (d) 5 to 4

Explanation:
1. Infrared needs a smaller energy gap than the 4 to 3 gap.
2. Gaps between consecutive levels shrink at higher n, so 5 to 4 has the smaller gap.

Teacher's Note:
1. Energy gaps between adjacent levels decrease as n increases.

Answer: (c) three

Explanation:
1. Energy of the level reached: \( -13.6+12.1=-1.5 \) eV, which is n = 3.
2. Lines from n = 3: 3 to 2, 3 to 1 and 2 to 1, so three lines.

Teacher's Note:
1. Number of lines from level n is n(n-1)/2.

Answer: (a) 79.0

Explanation:
1. The second electron is removed from He\( ^+ \) (Z = 2): \( 13.6 imes4=54.4 \) eV.
2. Total \( 24.6+54.4=79.0 \) eV.

Teacher's Note:
1. He+ is hydrogen-like with Z = 2.

Answer: (a) \( K_n \) independent of n, \( r_n\propto n \)

Explanation:
1. Centripetal condition: \( \frac{mv^2}{r}=\frac kr \), so \( mv^2=k \) is independent of n, hence K is the same for all n.
2. Quantisation \( mvr=\frac{nh}{2\pi} \) with constant v gives \( r\propto n \).

Teacher's Note:
1. A constant speed makes r proportional to n.

Answer: (a) \( \lambda_1=\lambda_2=4\lambda_3=9\lambda_4 \)

Explanation:
1. Neglecting the small nuclear-mass correction, \( \frac1\lambda\propto Z^2 \).
2. Z = 1, 1, 2, 3 give \( \lambda_1=\lambda_2=4\lambda_3=9\lambda_4 \).

Teacher's Note:
1. H and D have almost identical spectra.

Answer: (c) \( \frac1{n^3} \)

Explanation:
1. \( f\propto\frac1{(n-1)^2}-\frac1{n^2}=\frac{2n-1}{n^2(n-1)^2} \).
2. For large n this is about \( \frac{2n}{n^4}=\frac2{n^3} \), so \( f\propto\frac1{n^3} \) (equal to the orbital frequency, as the correspondence principle demands).

Teacher's Note:
1. For large n, the radiated frequency equals the classical orbital frequency.

Answer: (a) \( n_1=4,\ n_2=2 \)

Explanation:
1. \( T\propto n^3 \), so \( n_1^3=8n_2^3 \) gives \( n_1=2n_2 \).
2. Of the pairs, only 4 and 2 satisfy \( n_1=2n_2 \) (the pair 6 and 3 would also work but is not offered).

Teacher's Note:
1. Period varies as the cube of n.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For the first orbit n = 1, which is the smallest permitted value, so the minimum is \( h/2\pi \).
2. R is the quantisation rule and explains A.

Teacher's Note:
1. Angular momentum is quantised in integer multiples of h/2 pi.

Answer: (a) A-q, B-r, C-p,t, D-s

Explanation:
1. \( B\propto\frac fr \) with \( f\propto\frac{Z^2}{n^3} \) and \( r\propto\frac{n^2}{Z} \) gives \( \frac{Z^3}{n^5} \) for the magnetic field (A-q).
2. Potential energy magnitude \( \propto\frac{Z^2}{n^2} \) (B-r); angular speed and frequency \( \propto\frac{Z^2}{n^3} \) (C-p, t); speed \( \propto\frac Zn \) (D-s).

Teacher's Note:
1. Combine v ~ Z/n and r ~ n^2/Z.

Answer: (a) A-s, B-q, C-p, D-r

Explanation:
1. The series is named by the final level: 1 Lyman, 2 Balmer, 3 Paschen, 4 Brackett.
2. So 5 to 2 is Balmer, 8 to 4 Brackett, 3 to 1 Lyman, 4 to 3 Paschen.

Teacher's Note:
1. Lyman is UV, Balmer visible, the rest infrared.

Answer: (a) 3.4 eV and 6.65 Å

Explanation:
1. For a Bohr orbit \( K=-E=3.4 \) eV.
2. \( \lambda=\frac{h}{\sqrt{2mK}}=\frac{6.63\times10^{-34}}{\sqrt{2\times9.11\times10^{-31}\times3.4\times1.6\times10^{-19}}}\approx6.65\times10^{-10} \) m = 6.65 Å.

Teacher's Note:
1. Kinetic energy is the negative of total energy.

Answer: (b) \( 2.11\times10^{-34} \) J s

Explanation:
1. Energy -3.4 eV corresponds to n = 2 for hydrogen.
2. \( L=\frac{nh}{2\pi}=2\times1.055\times10^{-34}=2.11\times10^{-34} \) J s.

Teacher's Note:
1. E_n = -13.6/n^2 eV.

Answer: (a) 48.4 eV and 122.4 V

Explanation:
1. He\( ^+ \): ground level -54.4 eV, n = 3 level -6.04 eV, so excitation energy is 48.4 eV.
2. Li\( ^{2+} \) ground-state ionisation energy \( 13.6 imes9=122.4 \) eV, so 122.4 V.

Teacher's Note:
1. Use E_n = -13.6 Z^2/n^2 eV.