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konstantin123 [22]
2 years ago
10

calculate the frequency of the light emitted by a hydrogen atom during a transition of its electron from the n = 3 to n = 1 ener

gy level based on the bohr theory
Chemistry
1 answer:
MissTica2 years ago
3 0

Answer:

\nu=2.92\times 10^{15}\ Hz

Explanation:

The expression for the energy of an electron in the nth orbit is:-

E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

Given, n_i=3\ and\ n_f=1

\Delta E=2.18\times 10^{-18}(\frac{1}{3^2} - \dfrac{1}{1^2})\ J

\Delta E=-1.94\times 10^{-18}\ J  (negative sign indicates energy release)

Also, E=h\times \nu

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

1.94\times 10^{-18}=6.626\times 10^{-34}\times \nu

\nu=2.92\times 10^{15}\ Hz

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