Question

A hydrogen atom changes its state from n = 3 to n = 2. Due to recoil, the percentage change in the wave length of emitted light is approximately 1 × 10^–n . The value of n is__________. 

[Given Rhc = 13.6 eV, hc = 1242 eV nm, h = 6.6 × 10^–34 J s, mass of the hydrogen atom = 1.6 × 10^–27 kg]

Answer 1

Correct answer is :  7

\(\quad \Delta \mathrm{E}=13.6\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=1.9 \mathrm{eV}\)

\(\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}\)

\(\lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}}\)

\(P_{i}=P_{f}\)

\(0=-m v+\frac{h}{\lambda^{\prime}}\)

\(\Rightarrow \mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \lambda^{\prime}}\)

\(\Delta \mathrm{E}=\frac{1}{2} \mathrm{mv}^{2}+\frac{\mathrm{hc}}{\lambda^{\prime}}\)

\(=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{h}}{\mathrm{m} \lambda^{\prime}}\right)^{2}+\frac{\mathrm{hc}}{\lambda^{\prime}}\)

Now

\(\Delta \mathrm{E}=\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{\prime 2}}+\frac{\mathrm{hc}}{\lambda^{\prime}}\)

\(\lambda^{\prime 2} \Delta \mathrm{E}-\mathrm{hc} \lambda^{\prime}-\frac{\mathrm{h}^{2}}{2 \mathrm{~m}}=0\)

\(\lambda^{\prime}=\frac{\mathrm{hc} \pm \sqrt{\mathrm{h}^{2} \mathrm{c}^{2}+\frac{4 \Delta \mathrm{E} \mathrm{h}^{2}}{2 \mathrm{~m}}}}{2 \Delta \mathrm{E}}\)

\(\lambda^{\prime}=\frac{\mathrm{hc} \pm \mathrm{hc} \sqrt{1+\frac{2 \Delta \mathrm{E}}{\mathrm{mc}^{2}}}}{2 \Delta \mathrm{E}}\) 

\(\frac{\lambda^{\prime}}{\lambda}=\frac{1+\left(1+\frac{2 \Delta \mathrm{E}}{\mathrm{mc}^{2}}\right)^{\frac{1}{2}}}{2}=\frac{1+1+\frac{\Delta \mathrm{E}}{\mathrm{mc}^{2}}}{2}\)

\(\frac{\lambda^{\prime}}{\lambda}=1+\frac{\Delta \mathrm{E}}{2 \mathrm{mc}^{2}}\)

\(\frac{\lambda^{\prime}-\lambda}{\lambda}=\frac{\Delta \mathrm{E}}{2 \mathrm{mc}^{2}}=\frac{1.9 \times 1.6 \times 10^{-19}}{2 \times 1.67 \times 10^{-27} \times 9 \times 10^{16}}=10^{-9}\)

\(\therefore \%\ change \approx 10^{-7}\)