Question

What is the frequency of the photons emitted by hydrogen atoms when they undergo transitions from n = 5 to n = 2? Calculate in s-1 In which region of the electromagnetic spectrum does this radiation occur?

Answer 1

We know that,

1/λ=Rh [(1/nf^2)-(1/ni^2)]

Where Rh=Rydberg constant=1.0974x10^7 m^-1.

ni=initial state and nf=final state.

Given ni=5 and nf=2,

Then 1/λ=(1.0974x10^7 m^-1)[(1/2^2)-(1/5^2)]

1/λ=2.3x10^6 m^-1

λ=1/(2.3x10^6 m^-1)=4.339x10^-7 m.

Wavelength=433.9x10^-9 m=433.9 nm (since 1 nm=10^-9 m). the

We know that frequency, v=c/λ

Where c=3x10^8 m/s

Therefore v=(3x10^8 m/s)/(4.339x10^-7 m)

Frequency=6.9x10^14 s^-1.

Here wavelength is ~434 nm, so this radiation occurs in visible region of electromagnetic spectrum.

(Visible range:400-800 nm)

Please let me know if you have any doubt. Thanks.