Question

Determine the wavelength of the photon associated with the transition in which the electron in a hydrogen atom goes from n = 6 to n = 4 a photon with a wavelength of 2694 nm is released a photon with a wavelength of 2694 nm is absorbed a photon with a wavelength of 2625 nm is absorbed a photon with a wavelength of 2533 nm is released a photon with a wavelength of 2625 nm is released

Answer 1

Answer:

a photon with a wavelength of 2625 nm is released

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, $n_i=6\ and\ n_f=4$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (\frac{1}{6^2} - \dfrac{1}{4^2})}\ m$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179\left(|\frac{1}{36}-\frac{1}{16}\right)|}\ m$

$\lambda=\frac{19.878}{10^8\times \:2.179\left(|-\frac{5}{144}\right|)}\ m$

$\lambda=2.625\times 10^{-6}\ m$

1 m = 10⁻⁹ nm

$\lambda=2625\ nm$

From coming from higher energy level to lower, energy is released.

Hence, correct option is - a photon with a wavelength of 2625 nm is released