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:
Answers
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:
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: