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  • The wavelength of radiation emitted is , when an electron jumps from the third to the second orbit of hydr

The wavelength of radiation emitted is \mathrm{\lambda _{0}}, when an electron jumps from the third to the second orbit of hydrogen atom. For the electron Jump from the fourth to second orbit of the hydrogen atom, the wavelength of radiation emitted will be

Option: 1

\frac{16}{25} \lambda_0


Option: 2

\frac{20}{27} \lambda_0


Option: 3

\frac{27}{20} \lambda_0


Option: 4

\frac{25}{16} \lambda_0


Answers (1)

best_answer

\mathrm{\frac{1}{\lambda_0}=R\left(\frac{1}{22}-\frac{1}{32}\right)=R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5 R}{36}-1 }

\mathrm{\frac{1}{\lambda^{\prime}}=R\left(\frac{1}{22}-\frac{1}{42}\right)=R\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{3 }{4}\frac{R}{16}-(2) }

\mathrm{ \Rightarrow \lambda^{\prime}=\frac{20}{27} \, \lambda_0 }

Posted by

Ritika Jonwal

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