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  2. Help me understand! - Atoms And Nuclei - JEE Main-3
# Engineering

The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first exited state. Ratio of respective wavelengths, \lambda _{1}/\lambda _{2}, of the photons emitted in this process is :

 

  • Option 1)

    20/7

  • Option 2)

    27/5

  • Option 3)

    7/5

  • Option 4)

    9/7

 

Answers (1)
S solutionqc

\frac{1}{\lambda _{1}}=R\left ( \frac{1}{3^{2}}-\frac{1}{4^{2}} \right )=\frac{7R}{144}\cdots \cdots (1)

\frac{1}{\lambda _{2}}=R\left ( \frac{1}{2^{2}}-\frac{1}{3^{2}} \right )=\frac{5R}{36}\cdots \cdots (2)

\frac{(2)}{(1)}=\frac{\lambda _{1}}{\lambda _{2}}=\frac{5R}{36}\times \frac{144}{7R}=\frac{20}{7}


Option 1)

20/7

Option 2)

27/5

Option 3)

7/5

Option 4)

9/7

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