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  4. An electron in hydrogen atom first jumps from second excited state to first excited state and then from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons emitted in these two cases be, a, b and c respectively. Then :

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Q. An electron in hydrogen atom first jumps from second excited state to first excited state and then from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons emitted in these two cases be, $a, b$ and $c$ respectively. Then :

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Solution:

Wavelenths emitted in the two cases are
$\lambda_{21}=\frac{\lambda_{32}}{\lambda_{21}} \text { and } l_{32}=\frac{12431}{1.89\, e V}$
$a=\frac{\lambda_{32}}{\lambda_{21}}=\frac{10.2}{1.89}=\frac{27}{5}$
Momentum ratio of photons is
$b=\frac{\lambda / \lambda_{32}}{\lambda / \lambda_{21}}=\frac{5}{27}$
Energy ratio of photons is
$c=\frac{h c / \lambda_{32}}{h c / \lambda_{21}}=\frac{5}{27}=\frac{1}{a}$