Respuesta :
Answer:
a photon with a wavelength of 2625 nm is released
Explanation:
[tex]E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules[/tex]
For transitions:
[tex]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[/tex]
[tex]\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J[/tex]
Also, [tex]\Delta E=\frac {h\times c}{\lambda}[/tex]
Where,
h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
So,
[tex]\frac {h\times c}{\lambda}=2.179\times 10^{-18}(|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)\ J[/tex]
[tex]\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[/tex]
So,
[tex]\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[/tex]
Given, [tex]n_i=6\ and\ n_f=4[/tex]
[tex]\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[/tex]
[tex]\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179\left(|\frac{1}{36}-\frac{1}{16}\right)|}\ m[/tex]
[tex]\lambda=\frac{19.878}{10^8\times \:2.179\left(|-\frac{5}{144}\right|)}\ m[/tex]
[tex]\lambda=2.625\times 10^{-6}\ m[/tex]
1 m = 10⁻⁹ nm
[tex]\lambda=2625\ nm[/tex]
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
