Question

What are the frequency and wavelength of a photon emitted during a transition from $n=5\text{state to the}n=2$ state in the hydrogen atom?

Answer 1

Calculation for emitted energy

$E_n=-2.18\times {10}^{-18}\times \frac{z^2}{n^2}J$

$\Delta E=E_f-E_i=2.18\times {10}^{-18}[\frac{1}{n_f^2}-\frac{1}{n_f^2}]$

Where, $n_i\to \text{initial energy~(orbit=5)}$
$n_f\to \text{final energy~(orbit=2)}$

(This transition gives rise to a spectral line in the visible region of the Balmer series.)

$\Delta E=2.18\times {10}^{-18}[\frac{1}{5^2}-\frac{1}{2^2}]J$

$=2.18\times {10}^{-18}\left[\frac{4-25}{25\times 4}\right]J$

$=-2.18\times {10}^{-18}\times \frac{21}{100}J$

$\Delta E=-4.58\times {10}^{19}J$
It is the emission energy,now we have to find out correspondig frequency and wavelength

Calculation for frequency (v)

$\because \text{We know that}\Delta E=hv=\frac{hc}{\lambda }$

$\Rightarrow v=\frac{\Delta E}{h}$

Where,$h=\text{planck constant}=6.626\times {10}^{-34}Js$
$\Delta E=\text{emission energy}=4.58\times {10}^{-19}J$
$v=?\left(\text{Frequency}\right)$
Susbstituting the values in the equation, we get

$v=\frac{4.58\times {10}^{-19}J}{6.626\times {10}^{24}Js}=0.691\times {10}^{-19+34}$

$v=6.91\times {10}^{14}Hz\left(se{c}^{-1}\right)$

Calculation for wavelength $\left(\lambda \right)$

$\because \text{We know,}\Delta E=hv=\frac{hc}{\lambda }\{v=\frac{c}{\lambda }\}$
Where,$c\to \text{velocity of the light}=3\times {10}^{8}m{s}^{-1}$

$\lambda \to \text{wavelength of the photon}=?$

$\Rightarrow \lambda =\frac{hc}{\Delta E}=\frac{6.626\times {10}^{-34}Js\times 3\times {10}^{8}m{s}^{-1}}{4.58\times {10}^{-19}J}$

$\lambda =\frac{19.878\times {10}^{-34+8}}{4.58\times {10}^{-19}}m$

$\lambda =4.34\times {10}^{-34+8+19}=4.34\times {10}^{-7}m$

$\lambda =434\times {10}^{-9}=434nm$ $\{1nm={10}^{-9}m\}$

Final answer: $v=6.91\times {10}^{14}Hz$
$\lambda =434nm$