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  • 9E, 3E and E are the K.E of an , a proton and an alpha particle. What's the qualitative order of their Brogl

9E, 3E and E are the K.E of an \mathrm{e^{-}}, a proton and an alpha particle. What's the qualitative order of their Broglie wavelengths?

Option: 1

\mathrm{\lambda e>\lambda p=\lambda \alpha}


Option: 2

\mathrm{\lambda e>\lambda p>\lambda \alpha}


Option: 3

\mathrm{\lambda e<\lambda p<\lambda \alpha}


Option: 4

\lambda_\alpha<\lambda_e=\lambda p


Answers (1)

best_answer

As we know, deBroglie wavelength is :-

\mathrm{\lambda=\frac{h}{m v}=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}}

So, 

\mathrm{\lambda e=\frac{h}{\sqrt{2\left(m_p / 1800\right) \times 9 E}}}

\mathrm{\lambda p=\frac{h}{\sqrt{2 m_p \times 3 E}}}

\mathrm{\lambda \alpha=\frac{h}{\sqrt{2 \times 4m p \times E}}         

So, 

\mathrm{\lambda e>\lambda p=\lambda \alpha}

Option (1) is right.

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