Question
A hydrogen atom initially in the ground level absorbs a photon, which excites it to the $n = 4$ level. Determine the wavelength and frequency of photon.
✓
Answer
For ground level $, n_1 = 1$
Let $E_1$ be the energy of this level. It ls known that $E_1$ Is related with $n_1$ as:
$\text{E}_1=\frac{-13.6}{\text{n}^2_1}\text{ eV}$
$=\frac{-13.6}{1^2}=-13.6\text{ eV}$
The atom is excited to a higher level $, n_2 = 4.$
Let $E_2$ be the energy of this level.
$\therefore\ \text{E}_2=\frac{-13.6}{\text{n}^2_2}\text{ eV}$
$=\frac{-13.6}{4^2}=-\frac{13.6}{16}\text{ eV}$
The amount of energy absorbed by the photon is given as:
$E = E_2 - E_1$
$=\frac{-13.6}{16}-\Big(-\frac{13.6}{1}\Big)$
$=\frac{13.6\times15}{16}\text{ eV}$
$=\frac{13.6\times15}{16}\times1.6\times10^{-19}=2.04\times10^{-18}\text{ J}$
For a photon of wavelenqtha. the expression of energy Is written as:
$\text{E}=\frac{\text{hc}}{\lambda}$
Where,
$h =$ Planck's constant $= 6.6 x 10^{-34} Js$
$c =$ Speed of light $= 3 x 10^8m/s$
$\therefore\ \lambda=\frac{\text{hc}}{\text{E}}$
$=\frac{6.6\times10^{-34}\times3\times10^8}{2.04\times10^{-18}}$
$=9.7\times10^{-8}\text{ m}=97\text{ nm}$
And, frequency of a photon is given by the relation,
$\text{v}=\frac{\text{c}}{\lambda}$
$=\frac{3\times10^8}{9.710^{-8}}\approx3.1\times10^{15}\text{ Hz}$
Hence, the wavelength of the photon is $97 \ nm$ while the frequency is $3.1 x 10^{15} Hz.$
Let $E_1$ be the energy of this level. It ls known that $E_1$ Is related with $n_1$ as:
$\text{E}_1=\frac{-13.6}{\text{n}^2_1}\text{ eV}$
$=\frac{-13.6}{1^2}=-13.6\text{ eV}$
The atom is excited to a higher level $, n_2 = 4.$
Let $E_2$ be the energy of this level.
$\therefore\ \text{E}_2=\frac{-13.6}{\text{n}^2_2}\text{ eV}$
$=\frac{-13.6}{4^2}=-\frac{13.6}{16}\text{ eV}$
The amount of energy absorbed by the photon is given as:
$E = E_2 - E_1$
$=\frac{-13.6}{16}-\Big(-\frac{13.6}{1}\Big)$
$=\frac{13.6\times15}{16}\text{ eV}$
$=\frac{13.6\times15}{16}\times1.6\times10^{-19}=2.04\times10^{-18}\text{ J}$
For a photon of wavelenqtha. the expression of energy Is written as:
$\text{E}=\frac{\text{hc}}{\lambda}$
Where,
$h =$ Planck's constant $= 6.6 x 10^{-34} Js$
$c =$ Speed of light $= 3 x 10^8m/s$
$\therefore\ \lambda=\frac{\text{hc}}{\text{E}}$
$=\frac{6.6\times10^{-34}\times3\times10^8}{2.04\times10^{-18}}$
$=9.7\times10^{-8}\text{ m}=97\text{ nm}$
And, frequency of a photon is given by the relation,
$\text{v}=\frac{\text{c}}{\lambda}$
$=\frac{3\times10^8}{9.710^{-8}}\approx3.1\times10^{15}\text{ Hz}$
Hence, the wavelength of the photon is $97 \ nm$ while the frequency is $3.1 x 10^{15} Hz.$
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