JEE Physics Atoms MCQs Set 12

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

Explanation:
1. \( \frac1\lambda=RZ^2\left(1-\frac14\right)\Rightarrow\lambda\propto\frac1{Z^2} \) (the small mass difference between H and D is neglected).
2. Z = 1, 1, 2, 3, so \( \lambda_1=\lambda_2=4\lambda_3=9\lambda_4 \).

Teacher's Note:
1. Wavelength of a hydrogen-like transition varies inversely with the square of the atomic number.

Answer: (a) \( r_n\propto n \)

Explanation:
1. Force \( F=e\frac{dV}{dr}=\frac{eV_0}{r} \) provides the centripetal force: \( \frac{mv^2}{r}=\frac{eV_0}{r} \), so v is constant.
2. Quantisation \( mvr=\frac{nh}{2\pi} \) gives \( r\propto n \)

Teacher's Note:
1. A logarithmic potential gives a force inversely proportional to r, so the speed is independent of the orbit.

Answer: (b) 5/27

Explanation:
1. Lyman longest: \( \frac1{\lambda_L}=R\left(1-\frac14\right)=\frac{3R}4 \); Balmer longest: \( \frac1{\lambda_B}=R\left(\frac14-\frac19\right)=\frac{5R}{36} \).
2. \( \frac{\lambda_L}{\lambda_B}=\frac{5/36}{3/4}=\frac5{27} \)

Teacher's Note:
1. The longest wavelength of a series corresponds to the transition from the next level.

Answer: (d) \( 1.46\times10^6 \) m/s

Explanation:
1. \( v_n=\frac{2\pi KZe^2}{nh}=2.18\times10^6\times\frac Zn\ \text{m/s} \)
2. \( v_3=2.18\times10^6\times\frac23=1.46\times10^6\ \text{m/s} \)

Teacher's Note:
1. Orbital speed in Bohr's model is proportional to Z/n.

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

Explanation:
1. \( \nu\propto\frac1{(n-1)^2}-\frac1{n^2}=\frac{2n-1}{n^2(n-1)^2}\approx\frac2{n^3} \)

Teacher's Note:
1. For large n the difference of the two inverse squares is about 2/n cubed.

Answer: (d) 6

Explanation:
1. Number of lines \( =\frac{n(n-1)}2=\frac{4\times3}2=6 \): 4-3, 4-2, 4-1, 3-2, 3-1, 2-1.

Teacher's Note:
1. Count all downward transitions among the four levels.

Answer: (b) 4

Explanation:
1. Photon energy \( =13.6\left(1-\frac14\right)=10.2\ \text{eV} \).
2. Ionisation energy of \( Li^{2+} \) from level n: \( \frac{13.6\times9}{n^2}\le10.2\Rightarrow n^2\ge12 \).
3. The least integer is n = 4 (n = 3 would need 13.6 eV).

Teacher's Note:
1. The photon must carry at least the binding energy of the level.

Answer: (a) 20.4 eV

Explanation:
1. The neutron and the atom have almost equal masses. In a perfectly inelastic head-on collision half the kinetic energy is lost.
2. The atom needs at least 10.2 eV (n = 1 to n = 2), so \( \frac12K\ge10.2\Rightarrow K\ge20.4\ \text{eV} \)

Teacher's Note:
1. Only half of the neutron's kinetic energy is available for excitation when the masses are equal.

Answer: (b) Four times the de-Broglie wavelength of the electron in the ground state

Explanation:
1. \( 2\pi r_n=n\lambda_n\Rightarrow\lambda_n=\frac{2\pi r_1n^2}{n}=2\pi r_1n \), so \( \lambda_n\propto n \).

Teacher's Note:
1. The circumference of the n-th orbit holds n wavelengths, and the radius grows as n squared.

Answer: (c) Its kinetic energy increases but potential energy and total energy decrease

Explanation:
1. \( K=-E=\frac{13.6Z^2}{n^2} \) rises as n falls, \( U=2E \) and E become more negative.
2. So kinetic energy increases, while potential and total energy decrease.

Teacher's Note:
1. In an inner orbit the electron is faster and more tightly bound.

Answer: (b) \( n\left(\frac{hqB}{4\pi m}\right) \)

Explanation:
1. \( qvB=\frac{mv^2}{r}\Rightarrow v=\frac{qBr}m \); quantisation \( mvr=\frac{nh}{2\pi}\Rightarrow qBr^2=\frac{nh}{2\pi} \).
2. \( E=\frac12mv^2=\frac{q^2B^2r^2}{2m}=\frac{qB}{2m}\cdot\frac{nh}{2\pi}=n\frac{hqB}{4\pi m} \)

Teacher's Note:
1. Combine the circular motion condition with the angular momentum quantisation.

Answer: (c) 2.86

Explanation:
1. \( E=13.6\left(\frac14-\frac1{25}\right)=13.6\times0.21=2.86\ \text{eV} \)

Teacher's Note:
1. The absorbed energy equals the difference of the two level energies.

Answer: (b) 2

Explanation:
1. \( U=-2K \) in every Bohr orbit, so \( |U|/K=2 \).

Teacher's Note:
1. For a Coulomb orbit the potential energy is twice the total energy and minus twice the kinetic energy.