Question

When there occurs a transition from $n=3$ to $n=2$ state in the hydrogen, the atom recoils. If $E_1-E_2$ = $1.9eV$, then the ratio of $KE$ of recoil and the photon energy is of the order of

• A. ${10}^{{}_8}$
• B. ${10}^{-9}$
• C. ${10}^{-6}$
• D. ${10}^{-11}$

Answer 1

Answer: (B)

${10}^{-9}$

Due the transition from $n=3$ to $n=2$ state of hydrogen atom, a photon releases having an energy equal to that of transition energy. Also as the photon has its own momentum, thus the hydrogen atom will recoil to conserve the linear momentum of the system .

Let the wavelength of the photon be $\lambda .$

Thus energy of the photon is given by: $E\left(eV\right)=\frac{hc}{\lambda }=\frac{1242}{\lambda \left(nm\right)}$

$\therefore 1.9eV=\frac{1242}{\lambda \left(nm\right)}⟹\lambda =653.68nm$

Now momentum of the photon, $P_{photon}=\frac{h}{\lambda }$

Let recoil velocity of the atom be $v$.

Applying conservation of momentum: $m_{H}v=\frac{h}{\lambda }$

$\therefore (1.67\times {10}^{-27})v=\frac{6.626\times {10}^{-34}}{653.68\times {10}^{-9}}⟹v=0.607m/s$

Now recoil energy of the atom, $E_{Recoil}=\frac{1}{2}{m}_H{v}^2=\frac{1}{2}\times (1.67\times {10}^{-27})\times (0.607{)}^2=3.08\times {10}^{-28}J=1.922\times {10}^{-9}eV$

$\therefore \frac{E_{Recoil}}{E_{photon}}=\frac{1.922\times {10}^{-9}eV}{1.9eV}=1.012\times {10}^{-9}$