Question

A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n=4 level. Determine the wavelength and frequency of photon.

Answer 1

The correct answer is: Wavelength=97nm while Frequency=\(3.1×10^{15}Hz.\)
For ground level, \(n_1=1\)
Let \(E_1\) be the energy of this level. It is known that \(E_1\) is related with \(n_1\) as:
\(E_1=\frac{-13.6}{n^2_1}eV\)
\(=\frac{-13.6}{1^2}=-13.6eV\)
The atom is excited to a higher level, \(n_2=4\)
Let \(E_2\) be the energy of this level.
\(∴E_2=\frac{-13.6}{n^2_2}eV\)
\(=\frac{-13.6}{4^2}=(\frac{-13.6}{16})eV\)
The amount of energy absorbed by the photon is given as:
\(E=E_2-E_1\)
\(=\frac{-13.6}{16}-(\frac{-13.6}{1})\)
\(=\frac{13.6×15}{16}eV\)
\(\frac{13.6×15}{16}×1.6×10^{-19}=2.04×10^{-18}J\)
For a photon of wavelength \(λ\), the expression of energy is written as:
\(E=\frac{hc}{λ}\)
Where,
h=planck's constant\(=6.6×10^{-34}Js\)
c=speed of light\(=3×10^8m/s\)
\(∴λ=\frac{hc}{E}\)
\(=\frac{6.6×10^{-34}×3×10^8}{2.04×10^{-18}}\)
\(=9.7×10^{-8}m=97nm\)
And,frequency of a photon is given by the relation,
\(v=\frac{c}{λ}\)
\(=\frac{3×10^8}{9.7×10^{-8}}≈3.1×10^{15}Hz\)
Hence,the wavelength of the photon is 97nm while the frequency is \(3.1×10^{15}Hz.\)