# Physics Class 12 NCERT Solutions: Chapter 12 Atoms Part 6 (For CBSE, ICSE, IAS, NET, NRA 2022)

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Q: 13. Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level to level . For large , show that this frequency equals the classical frequency of revolution of the electron in the orbit.

It is given that a hydrogen atom de-excites from an upper level (n) to a lower level .

We have the relation for energy of radiation at level n as:

Where,

Plank՚s constant

Mass of hydrogen atom

charge on an electron

Permittivity of free space

Now, the relation for energy of radiation at level is given as:

Where,

Energy (E) released as a result of de-excitation:

Where,

Putting values from equations (i) and (ii) in equation (iii) , we get:

For large n, we can write and .

Classical relation of frequency of revolution of an electron is given as:

Where,

Velocity of the electron in the nth orbit is given as:

And, radius of the nth orbit is given as:

Putting the values of equations (vi) and (vii) in equation (v) , we get:

Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.