A hydrogen atom initially in the ground level absorbs a photon,

Solution:

For ground level, n1 = 1

Let E1 be the energy of this level. It is known that E1 is related with n1 as:

$E_{1}=\frac{-13.6}{n_{1}{ }^{2}} \mathrm{eV}$

$=\frac{-13.6}{1^{2}}=-13.6 \mathrm{eV}$

The atom is excited to a higher level, n2 = 4.

Let E2 be the energy of this level.

$\therefore E_{2}=\frac{-13.6}{n_{2}{ }^{2}} \mathrm{eV}$

$=\frac{-13.6}{4^{2}}=-\frac{13.6}{16} \mathrm{eV}$

The amount of energy absorbed by the photon is given as:

$E=E_{2}-E_{1}$

$=\frac{-13.6}{16}-\left(-\frac{13.6}{1}\right)$

$=\frac{13.6 \times 15}{16} \mathrm{eV}$

$=\frac{13.6 \times 15}{16} \times 1.6 \times 10^{-19}=2.04 \times 10^{-18} \mathrm{~J}$

For a photon of wavelengthλ, the expression of energy is written as:

$E=\frac{h c}{\lambda}$

Where,

h = Planck’s constant = 6.6 × 10−34 Js

c = Speed of light = 3 × 108 m/s

$\therefore \lambda=\frac{h c}{E}$

$=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{2.04 \times 10^{-18}}$

$=9.7 \times 10^{-8} \mathrm{~m}=97 \mathrm{~nm}$

And, frequency of a photon is given by the relation,

$v=\frac{c}{\lambda}$

$=\frac{3 \times 10^{8}}{9.7 \times 10^{-8}} \approx 3.1 \times 10^{15} \mathrm{~Hz}$

Hence, the wavelength of the photon is $97 \mathrm{~nm}$ while the frequency is $3.1 \times 10^{15} \mathrm{~Hz}$.