Question

In a hydrogen atom, in transition of electron a photon of energy 2.55 eV is emitted, then the change in wavelength of the electron is:

• A. $6.64\stackrel{o}{A}$
• B. $7.01\stackrel{o}{A}$
• C. $7.45\stackrel{o}{A}$
• D. $8.01\stackrel{o}{A}$

Answer 1

The correct option is A $6.64\stackrel{o}{A}$

The energy of the photon is $\Delta E=2.55eV=2.55\times 1.602\times {10}^{-19}J$.
The wavelength of the photon of light is $\lambda =\frac{hc}{\Delta E}=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{2.55\times 1.602\times {10}^{-19}m}=4861A^0$
This corresponds to the second line of Balmar series. The transition is $n_2=4\to n_1=2$
The change in the wavelength of the electron is $\Delta \lambda =\frac{h}{m{u}_4}-\frac{h}{m{u}_2}=\frac{h}{m}(u_4-u_2)=\frac{6.626\times {10}^{-34}}{9.1085\times {10}^{-31}}\left(\frac{4-2}{2.19\times {10}^m/s}\right)=6.64A^0$.