# 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.

Answer:

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,

Frequency of radiation at level

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,

Frequency of radiation level

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

Where,

Frequency of radiation emitted

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.