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  • Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength <img alt="975 {\AA}^{\circ}" src="https://entrancecorner.oncodecogs.com/gif.latex?975%20%7B%5CAA%7D%5E

Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 975 {\AA}^{\circ} The longest wavelength amongst them inthe emission spectrum was found 18.807 \times 10^{-y} \mathrm{~m} \text {. }. Then value of y is.

(Given : Ionisation energy for hydrogen atom as 13.6 eV.)

Option: 1

4


Option: 2

5


Option: 3

6


Option: 4

7


Answers (1)

best_answer

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

Energy of incident photon of wavelength

\mathrm{\begin{aligned} & \lambda=975 \AA=975 \times 10^{-10} \mathrm{~m} \\ & E=\frac{h c}{\lambda} \\ & =12.75 \mathrm{eV} \end{aligned}}

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{\mathrm{n}^{\text {th }}} orbit.

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

This gives 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 → 3)for which the energy difference is minimum

\mathrm{\begin{aligned} & E 4_3(-1.511)_{\min } \\ & =1.511-0.85=0.661 \mathrm{eV} \\ & =0.661 \times 1.6 \times 10^{-19} \mathrm{Joule} \\ & \quad \therefore \lambda \frac{\mathrm{hc}}{E_{\min } \frac{6.63 \times 10^{-34 \times 3 \times 10^8}}{0.661 \times 1.6 \times 10^{-19}}} \max \quad=18.807 \times 10^{-7} \mathrm{~m} \end{aligned}}

 

Posted by

rishi.raj

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