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Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength $975 \AA$ . The longest wavelength amongst them in the emission spectrum was found $18.807 \times 10^{-\mathrm{y}} \mathrm{m}$ . Value of $\mathrm{y}$ is. (Given : Ionisation energy for hydrogen atom as $13.6 \, \, \mathrm{eV}$ .)
Option: 1
5

Option: 2
6

Option: 3
7

Option: 4
8

Answers (1)

best_answer

$E_n=-\frac{R h c}{n^2}$

Energy of incident photon of wavelength

$\lambda=975 \stackrel{0}{\AA}=975 \times 10^{-10} \mathrm{~m}$

$\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$

When the incident photon of this energy is absorbed by hydrogen atom.

Let its ground state electron occupy $(n-1)$ the excited state or $\mathrm{n}^{\text {th }}$ orbit.

$=12.75 \mathrm{eV}=13.6 \mathrm{eV}\left(1-\frac{1}{\mathrm{n}^2}\right)$

This gives $\mathrm{n}=4$
That is the electron is excited to III excited state. The emission spectrum will contain the transitions. The longest wavelength emitted corresponds to transition $(4 \rightarrow 3)$ for which the energy difference is minimum

$\mathrm{E}_{\min }=\mathrm{E}_4-\mathrm{E}_3=-0.85-(-1.511) \mathrm{eV}$

$=1.511-0.85=0.661 \mathrm{eV}$

$=0.661 \times 1.6 \times 10^{-19} \text { Joule }$

$\therefore \lambda_{\max }=\frac{\mathrm{hc}}{\mathrm{E}_{\text {min }}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{0.661 \times 1.6 \times 10^{-19}}$

$=18.807 \times 10^{-7} \mathrm{~m}=18807\, \, \AA$

Posted by

himanshu.meshram

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