Question

The ratio of radius of two different orbits in a h atom is 4:9, then the ratio of the frequency of revolution of electron in these orbits is:

Answer 1

Hey Dear,

◆ Answer -

f/f' = 27/8

● Explanation -

Radius of nth Bohr orbit is directly proportional to square of orbit number.

Rn ∝ n^2

As ratio of orbits is given -

R/R' = (n/n')^2

4/9 = (n/n')^2

n/n' = 2/3

Frequency of revolution of electron in nth Bohr orbit is inversely proportional to cube of orbit number.

f ∝ 1/n^3

Therefore, ratio of frequencies of revolution is -

f/f' = (n'/n)^3

f/f' = (3/2)^3

f/f' = 27/8

Hence, ratio of frequencies of revolution of electrons is 27/8.

Hope this helped you.

Answer 2

Answer:

Hey Dear,

◆ Answer -

f/f' = 27/8

● Explanation -

Radius of nth Bohr orbit is directly proportional to square of orbit number.

Rn ∝ n^2

As ratio of orbits is given -

R/R' = (n/n')^2

4/9 = (n/n')^2

n/n' = 2/3

Frequency of revolution of electron in nth Bohr orbit is inversely proportional to cube of orbit number.

f ∝ 1/n^3

Therefore, ratio of frequencies of revolution is -

f/f' = (n'/n)^3

f/f' = (3/2)^3

f/f' = 27/8

Hence, ratio of frequencies of revolution of electrons is 27/8.

Hope this helped you.

Explanation: