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  • The radiation, emitted when an electron jumps from n=4 to n=3 in a Lithium atom, falls on a metal surface to produce photoelectron. wavelength of radiation falling on the metal surface is n

The radiation, emitted when an electron jumps from n=4 to n=3 in a Lithium atom, falls on a metal surface to produce photoelectron. wavelength of radiation falling on the metal surface is
 

Option: 1

2.08 \times 10^{-7} \mathrm{~m}


Option: 2

2.08 \times 10^{-8} \mathrm{~m}


Option: 3

3.08 \times 10^{-7} \mathrm{~m}


Option: 4

2.08 \times 10^{-9} \mathrm{~m}


Answers (1)

best_answer

The wavelength of emitted radiation is given as
(a)  \frac{1}{\lambda}=z^2 R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)

For the Lithium atom, \quad \mathrm{Z}=3

\begin{aligned} & \therefore \frac{1}{\lambda}=3^2 \mathrm{R}\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=9 \mathrm{R} \times \frac{7}{144} \\ & \lambda=2.08 \times 10^{-7} \mathrm{~m} \end{aligned}

 

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Gunjita

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