Login/Sign Up
Home » 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.
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.
CBSE | Class 12 | NCERT | Nuclei | Physics
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 –

We know that for ground level, n1 = 1

Suppose E1  is the energy of the level n1. So as we know E1 is related with n1 as –

    \[{{E}_{1}}=\frac{13.6}{{{n}_{1}}^{2}}eV=-13.6eV\]

As the atom is excited to a higher level given by n2= 4. Again, let E2 be the energy corresponding to this level. Then, similarly –

    \[{{E}_{2}}=\frac{13.6}{{{n}_{2}}^{2}}eV=\frac{-13.6}{{{4}^{2}}}eV\]

    \[{{E}_{2}}=\frac{-13.6}{16}eV\]

So, the amount of energy absorbed by photon is given by –

E = E2βˆ’ E1

    \[\therefore E=\frac{-13.6}{16}-\left( -13.6 \right)eV\]

    \[E=\frac{-13.6\times 15}{16}\times 1.6\times {{10}^{-19}}J\]

    \[E=2.04\times {{10}^{-18}}J\]

We know that for a photon of wavelength Ξ», energy is given by the relation –

    \[\lambda =\frac{hc}{E}\]

Where,                                                        

h is Planck’s constant, given = 6.6 Γ— 10βˆ’34 Js

c is the speed of light, given = 3 Γ— 108 m/s

So, by substituting value in equation we get –

    \[\lambda =\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2.04\times {{10}^{-18}}}m\]

    \[\lambda =97nm\]

We know that the wavelength and the frequency are related to each other by the relation –

    \[\nu =\frac{c}{\lambda }\]

    \[\therefore \nu =\frac{3\times {{10}^{8}}}{97\times {{10}^{-9}}}Hz\]

    \[\nu =3.1\times {{10}^{15}}Hz\]

Therefore, 97nm is the wavelength and 3.1Γ—1015Hz is the frequency of the photon.

i

More Material