Question

ratio of frequency of revolution of electron in the first excited state of He + n the third excited state of H

Answer 1

hey here is ur answer...

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Answer 2

Hey dear,

● Answer -
f1 : f2 = 32 : 1

● Explaination -
For electron in first excited state of He atom, Z1 = 2 & n1 = 2.

For electron in third excited state of H atom, Z2 = 1 & n2 = 4.

Frequency of revolution of electron is -
f ∝ Z^2 / n^3

Ratio of frequencies -
f1/f2 = (Z1/Z2)^2 × (n2/n1)^3
f1/f2 = (2/1)^2 × (4/2)^3
f1/f2 = 4 × 8
f1/f2 = 32

Therefore, ratio of frequencies is 32:1.

Hope this helps you...