Question

What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom ?

• A. $4.35\times {10}^{16}Hz$, 691 nm
• B. $6.91\times {10}^{14}Hz$, 434 nm
• C. $4.35\times {10}^{15}Hz$, 450 nm
• D. $6.91\times {10}^{13}Hz$, 400 nm

Answer 1

The correct option is B $6.91\times {10}^{14}Hz$, 434 nm

Since ni=5 and nf=2, this transition gives rise to a spectral line in the visible region of the Balmer series. From equation.
ΔE=2.18×10−18J[152−122]
=−4.58×10−19J
Negative sign implies that it is an emission energy.
The frequency of the photon (taking energy in terms of magnitude) is given by
v=ΔEh
=4.58×10−19J6.626×10−34Js
⇒6.91×1014 Hz
The wavelength of photon emitted (λ)=cv
=3.0×108 ms−16.91×1014Hz
=4.34×10−7m or 434 nm