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  • What is the wavelength of light for the least energetic photon emitted in the Lyman series of the hydrogen atom spectrum? (Take hc = 1240 eV nm)Option:

What is the wavelength of light for the least energetic photon emitted in the Lyman series of the hydrogen atom spectrum?

(Take hc = 1240 eV nm)

Option: 1

82 nm


Option: 2

102 nm


Option: 3

122 nm 


Option: 4

150 nm


Answers (1)

best_answer

For any series, the transition that produces the least energetic photon is the transition between the home-base level that defines the series and the level immediately above it.

For the Lyman series, the home-base level is at n = 1. Thus, the transition that produces the least energetic photon is the transition from the n = 2 level to the n = 1 level.

As, \mathrm{E_n=-\frac{13.6}{n^2} e V}

\begin{aligned} & \therefore \quad \mathrm{E_1}=-\frac{13.6}{1^2} \mathrm{eV}=-13.6 \mathrm{eV} \\ \\& \text { and } \mathrm{E}_2=-\frac{13.6}{2^2} \mathrm{eV}=-3.4 \mathrm{eV} \end{aligned}

The corresponding energy difference is \mathrm{\Delta \mathrm{E}=\mathrm{E}_2-\mathrm{E}_1=-3.4 \mathrm{eV}-(-13.6 \mathrm{eV})=10.2 \mathrm{eV}}

Wavelength of emitted photon is \lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}}=\frac{1240 \mathrm{eV} \mathrm{nm}}{10.2 \mathrm{eV}}=122 \mathrm{~nm}

Posted by

Devendra Khairwa

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