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Question: A proton, a neutron, an electron and an \(\alpha\) particle have same energy. Then their de Broglie ...

A proton, a neutron, an electron and an α\alpha particle have same energy. Then their de Broglie wavelengths compare as

A

λp=λn>λe>λα\lambda_{p} = \lambda_{n} > \lambda_{e} > \lambda_{\alpha}

B

λα<λp=λn<λe\lambda_{\alpha} < \lambda_{p} = \lambda_{n} < \lambda_{e}

C

λe<λp=λn>λα\lambda_{e} < \lambda_{p} = \lambda_{n} > \lambda_{\alpha}

D

λe=λp=λn=λα\lambda_{e} = \lambda_{p} = \lambda_{n} = \lambda_{\alpha}

Answer

λα<λp=λn<λe\lambda_{\alpha} < \lambda_{p} = \lambda_{n} < \lambda_{e}

Explanation

Solution

: kinetic energy of particle, K=12mv2K = \frac{1}{2}mv^{2}

Or mv=2mKmv = \sqrt{2mK}

De Broglie wavelength,

λ=hmv=h2mK\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mK}}

For the given value of K,λ∝1m\lambda \propto \frac{1}{\sqrt{m}}

∴λp:λn:λe:λα=1mp:1mn:1me:1mα\therefore\lambda_{p}:\lambda_{n}:\lambda_{e}:\lambda_{\alpha} = \frac{1}{\sqrt{m_{p}}}:\frac{1}{\sqrt{m_{n}}}:\frac{1}{\sqrt{m_{e}}}:\frac{1}{\sqrt{m_{\alpha}}}

Since mp=mn,henceλp=λnm_{p} = m_{n},hence\lambda_{p} = \lambda_{n}

As mα>mp,thereforeλα<λpm_{\alpha} > m_{p},therefore\lambda_{\alpha} < \lambda_{p}

As me<mn,henceλe>λnm_{e} < m_{n},hence\lambda_{e} > \lambda_{n}

Hence λα<λp=λn<λe\lambda_{\alpha} < \lambda_{p} = \lambda_{n} < \lambda_{e}