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DaniilM [7]
3 years ago
8

10. A triply ionized beryllium atom is in the ground state. It absorbs energy and makes a transition to the n = 5 excited state.

The ion returns to the ground state by emitting FOUR photons ONLY. What is the wavelength of the second highest energy photon?
Physics
1 answer:
Xelga [282]3 years ago
6 0

Answer:

\lambda=1282\ nm

Explanation:

E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

Also, \Delta E=\frac {h\times c}{\lambda}

Where,  

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

c is the speed of light having value 3\times 10^8\ m/s

So,  

\frac {h\times c}{\lambda}=2.179\times 10^{-18}(|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)\ J

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m

So,  

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m

Given, n_i=5\ and\ n_f=3

\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (\frac{1}{5^2} - \dfrac{1}{3^2})}\ m

\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179\left(|\frac{1}{25}-\frac{1}{9}\right)|}\ m

\lambda=\frac{19.878}{10^8\times \:2.179\left(|-\frac{16}{225}\right|)}\ m

\lambda= 1.2828\times10^{-6}

1 m = 10⁻⁹ nm

\lambda=1282\ nm

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