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Physics Question on Atoms

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

The correct answer is: Wavelength=97nm while Frequency=3.1×1015Hz.3.1×10^{15}Hz.
For ground level, n1=1n_1=1
Let E1E_1 be the energy of this level. It is known that E1E_1 is related with n1n_1 as:
E1=−13.6n12eVE_1=\frac{-13.6}{n^2_1}eV
=−13.612=−13.6eV=\frac{-13.6}{1^2}=-13.6eV
The atom is excited to a higher level, n2=4n_2=4
Let E2E_2 be the energy of this level.
∴E2=−13.6n22eV∴E_2=\frac{-13.6}{n^2_2}eV
=−13.642=(−13.616)eV=\frac{-13.6}{4^2}=(\frac{-13.6}{16})eV
The amount of energy absorbed by the photon is given as:
E=E2−E1E=E_2-E_1
=−13.616−(−13.61)=\frac{-13.6}{16}-(\frac{-13.6}{1})
=13.6×1516eV=\frac{13.6×15}{16}eV
13.6×1516×1.6×10−19=2.04×10−18J\frac{13.6×15}{16}×1.6×10^{-19}=2.04×10^{-18}J
For a photon of wavelength λλ, the expression of energy is written as:
E=hcλE=\frac{hc}{λ}
Where,
h=planck's constant=6.6×10−34Js=6.6×10^{-34}Js
c=speed of light=3×108m/s=3×10^8m/s
∴λ=hcE∴λ=\frac{hc}{E}
=6.6×10−34×3×1082.04×10−18=\frac{6.6×10^{-34}×3×10^8}{2.04×10^{-18}}
=9.7×10−8m=97nm=9.7×10^{-8}m=97nm
And,frequency of a photon is given by the relation,
v=cλv=\frac{c}{λ}
=3×1089.7×10−8≈3.1×1015Hz=\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 3.1×1015Hz.3.1×10^{15}Hz.