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Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65\AA). The de-Broglie wavelength of this electron is:

 

  • Option 1)

    3.5 \AA

  • Option 2)

    6.6 \AA

  • Option 3)

    12.9 \AA

     

  • Option 4)

    9.7 \AA

 

Answers (1)

best_answer

2\pi r_{n}=n\lambda _{n}

\lambda _{3}=\frac{2\pi (4.65\times 10^{-10})}{3}

\lambda _{3}=9.7\AA


Option 1)

3.5 \AA

Option 2)

6.6 \AA

Option 3)

12.9 \AA

 

Option 4)

9.7 \AA

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