Neet Solved Paper 2013 Question 38

Options:

A) $ {\lambda _p}\propto \lambda _e^{2} $

B) $ {\lambda _p}\propto {\lambda _e} $

C) $ {\lambda _p}\propto \sqrt{{\lambda _e}} $

D) $ {\lambda _p}\propto \frac{1}{\sqrt{{\lambda _e}}} $

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Answer:

Correct Answer: A

Solution:

  • [a] Wavelength of electron, $ {\lambda_e}=\frac{h}{\sqrt{2mE}} $ and proton $ {\lambda_p}=\frac{hc}{E} $
    $ \Rightarrow $ $ \lambda _e^{2}=\frac{h^{2}}{2mE} $ or $ E=\frac{hc}{{\lambda_p}} $
    $ \therefore $ $ \lambda _e^{2}=\frac{h^{2}}{2m.\frac{hc}{{\lambda_p}}}\Rightarrow \lambda _e^{2}=\frac{h _2}{2mhc}{\lambda_p} $
    $ \Rightarrow $ $ \lambda _e^{2}\propto {\lambda_p} $