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, n1 = 1
Let E1 be the energy of this level. It ls known that E1 Is related with n1 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, n2 = 4.
Let E2 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 = E2 - E1
$=\frac{-13.6}{16}-\Big(-\frac{13.6}{1}\Big)$
$=\frac{13.6\times15}{16}\text{ eV}$
$=\frac{13.6\times15}{16}\times1.6\times10^{-19}=2.04\times10^{-18}\text{ J}$
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 108 m/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 1015 Hz.
Let E1 be the energy of this level. It ls known that E1 Is related with n1 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, n2 = 4.
Let E2 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 = E2 - E1
$=\frac{-13.6}{16}-\Big(-\frac{13.6}{1}\Big)$
$=\frac{13.6\times15}{16}\text{ eV}$
$=\frac{13.6\times15}{16}\times1.6\times10^{-19}=2.04\times10^{-18}\text{ J}$
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 108 m/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 1015 Hz.
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