JEE Physics Atoms MCQs Set 10

Answer: (a) 94.1 nm

Explanation:
1. Energy of the photon: \( E=13.6\left(1-\frac1{16}\right)=12.75\ \text{eV} \).
2. \( \lambda=\frac{hc}{E}=\frac{4\times10^{-15}\times3\times10^{8}}{12.75}=9.41\times10^{-8}\ \text{m} \) = 94.1 nm.

Teacher's Note:
1. Use the energy level formula E_n = -13.6/n² eV for hydrogen.
2. hc/E gives the wavelength directly once h is given in eV s.

Answer: (a) 27

Explanation:
1. Second excited state is n = 3 and first excited state is n = 2: \( \frac1{\lambda_0}=R\left(\frac14-\frac19\right)=\frac{5R}{36} \).
2. Third excited state is n = 4 and the second orbit is n = 2: \( \frac1{\lambda'}=R\left(\frac14-\frac1{16}\right)=\frac{3R}{16} \).
3. \( \frac{\lambda'}{\lambda_0}=\frac{16}{3}\times\frac{5}{36}=\frac{20}{27} \), so x = 27.

Teacher's Note:
1. The ground state is n = 1, so the first excited state is n = 2 and the nth excited state is n + 1.
2. Wavelength is inversely proportional to the energy difference.

Answer: (d) D

Explanation:
1. Photon energy: \( E=\frac{hc}{\lambda}=\frac{6.62\times10^{-34}\times3\times10^{8}}{124.1\times10^{-9}}=1.6\times10^{-18}\ \text{J}=10\ \text{eV} \).
2. Energy differences: A = 2.2 eV, B = 5.2 eV, C = 3.0 eV, D = 10.0 eV.
3. Only transition D, from level 1 (0.0 eV) to level 4 (-10.0 eV), releases 10 eV.

Teacher's Note:
1. Convert the wavelength to energy first and then compare with the differences between levels.
2. hc is about 1240 eV nm, so 124 nm corresponds to 10 eV.

Answer: (d) \( 8.4\times10^{6} \)

Explanation:
1. Speed in the nth Bohr orbit: \( v_n\propto\frac1n \).
2. \( v_3=3.6\times10^6\times\frac73=8.4\times10^6\ \text{m/s} \)

Teacher's Note:
1. In Bohr's model the orbital speed falls as 1/n while the radius grows as n².
2. A lower orbit always means a faster electron.

Answer: (a) 27

Explanation:
1. \( \frac1{\lambda_1}=R\left(1-\frac19\right)=\frac{8R}9,\qquad\frac1{\lambda_2}=R\left(1-\frac14\right)=\frac{3R}4 \).
2. \( \frac{\lambda_1}{\lambda_2}=\frac{3/4}{8/9}=\frac{27}{32} \), so x = 27.

Teacher's Note:
1. A larger energy gap means a shorter wavelength, so lambda1 must be smaller than lambda2.
2. Use the Rydberg formula for each transition separately.

Answer: (b) 2.25 R

Explanation:
1. Orbit radius: \( r_n\propto n^2 \).
2. \( \frac{r_3}{r_2}=\frac{9}{4}\Rightarrow r_3=2.25R \)

Teacher's Note:
1. The radius is proportional to the square of the orbit number, n².
2. The first Bohr radius is about 0.53 Å.

Answer: (a) 136

Explanation:
1. Li²⁺ is hydrogen-like with Z = 3. The second excited state is n = 3.
2. Energy needed to ionise from level n: \( E=13.6\,\frac{Z^2}{n^2}=13.6\times\frac{9}{9}=13.6\ \text{eV}=136\times10^{-1}\ \text{eV} \)

Teacher's Note:
1. The ionisation energy of a hydrogen-like ion from level n is 13.6 Z²/n² eV.
2. Count excited states from n = 2 (first excited), so the second excited state is n = 3.

Answer: (a) 828

Explanation:
1. Energy of the level reached: \( -13.6+12.75=-0.85\ \text{eV}=-\frac{13.6}{n^2}\Rightarrow n=4 \).
2. Angular momentum: \( L=\frac{nh}{2\pi}=\frac{4\times4.14\times10^{-15}}{2\pi}=\frac{8.28\times10^{-15}}{\pi}=\frac{828}{\pi}\times10^{-17}\ \text{eV s} \).

Teacher's Note:
1. First find the level n from the final energy, then apply the quantisation rule L = n h / 2 pi.
2. The excited level is n = 4 because 13.6 x (1 - 1/16) = 12.75 eV.

Answer: (d) 40.8 eV

Explanation:
1. \( \Delta E=Rch\,Z^2\left(\frac1{n_f^2}-\frac1{n_i^2}\right)=13.6\times16\times\left(\frac14-\frac1{16}\right)=13.6\times16\times\frac3{16}=40.8\ \text{eV} \)

Teacher's Note:
1. The energy levels of a hydrogen-like atom scale with Z².
2. Compute 1/4 - 1/16 = 3/16 first; it cancels with the factor 16.