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

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

Dual Nature of Radiation and Matter

Solution:

Kinetic energy of particle, $K = \frac{1}{2} mv^2$
or $ mv = \sqrt {2mK}$
de Broglie wavelength,
$\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mK}}$
For the given value of $K, \lambda \propto \frac{1}{\sqrt{m}}$
$\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 $m_p = m_n$, hence $\lambda_p = \lambda_n$
As $m_\alpha > m_p$, therefore $\lambda _\alpha < \lambda_p$
As $m_e < m_n$, therefore $\lambda_e > \lambda_n$
Hence $\lambda_\alpha < \lambda_p = \lambda_n < \lambda_e$