Question

In a mixture of $H-H{e}^{+}$ gas, H atom and $H{e}^{+}$ ions are excited to their respective first excited states. Subsequently, H atoms transfer its excitation energy to $H{e}^{+}$ ions by collision.

If Bohr model is applicable then the wavelength of the light emitted in the visible region by $H{e}^{+}$ ions after collisions with $H{e}^{+}$ ions are:

• A. 400 A
• B. 344 A
• C. 244 A
• D. 375 A

Answer 1

The correct option is C 244 A

The energy difference between ground state and first excited state for H atom is
$E=21.79\times {10}^{-19}[\frac{1}{1^2}-\frac{1}{2^2}]$......(1)
The energy difference between ground state and first excited state for He+ ion is
$E^{\prime }=21.79\times {10}^{-19}\times 2^2[\frac{1}{1^2}-\frac{1}{2^2}]$......(2)
Total energy difference is obtained by adding (1) and (2)
$E"=E+E^{\prime }=21.79\times {10}^{-19}\times 3.75J/atom$
Wavelength of light emitted is $\lambda =\frac{hc}{E"}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{21.79\times {10}^{-19}\times 3.75}=2.44\times {10}^{-8}m=244A$.