Ncert Exercises & Solutions

12.6 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.

Hint: Energy recieved by the hydrogen atom equals to the energy difference between the levels.
Step 1: Find the energy of the atom in the ground level.
For ground level,

$n_{1} = 1$

Let

$E_{1}$ be the energy of this level.

It is known that

$E_{1}$ is related with

$n_{1}$ as:

$\begin{aligned} E_{1} & = \frac{- 13.6}{n_{1}^{2}} eV \\ = \frac{- 13.6}{1^{2}} = - 13.6 eV \end{aligned}$

Step 2: Find the energy of the atom in the excited level.
The atom is excited to a higher level,

$n_{2}$ = 4.

Let

$E_{2}$ be the energy of this level.

$\begin{array}{l} = \frac{- 13.6}{16} - (- \frac{13.6}{1}) \\ = \frac{13.6 \times 15}{16} eV \\ = \frac{13.6 \times 15}{16} \times 1.6 \times 10^{- 19} = 2.04 \times 10^{- 18} J \end{array}$

Step 3: Find the wavelength and frequency of the photon.
For a photon of wavelength

$λ$ , the expression of energy is written as:

$E = \frac{h c}{λ}$

where,

h=planck's constant

$= 6.62 \times 10^{- 34} Js a n d c = S p e e d o f l i g h t = 3 \times 10^{8} m / s$

$∴ λ = \frac{h c}{E}$

$= \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{2.04 \times 10^{- 18}}$

$= 9.74 \times 10^{- 8} m = 97.4 n m$

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

$\nu=\frac{c}{\lambda}=\frac{3 \times 10^{8}}{9.74 \times 10^{-8}} \approx 3.1 \times 10^{15} \mathrm{~Hz}$

Hence, the wavelength of the photone is 97.4 nm while the frequency is

$3.1 \times 10^{15} H z .$

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