JEE MAIN - Physics (2022 - 25th June Evening Shift - No. 16)
A proton, a neutron, an electron and an $$\alpha$$ particle have same energy. If $$\lambda$$p, $$\lambda$$n, $$\lambda$$e and $$\lambda$$a are the de Broglie's wavelengths of proton, neutron, electron and $$\alpha$$ particle respectively, then choose the correct relation from the following :
$$\lambda$$p = $$\lambda$$n > $$\lambda$$e > $$\lambda$$a
$$\lambda$$a < $$\lambda$$n < $$\lambda$$p < $$\lambda$$e
$$\lambda$$e < $$\lambda$$p = $$\lambda$$n > $$\lambda$$a
$$\lambda$$e = $$\lambda$$p = $$\lambda$$n = $$\lambda$$a
Explanation
$$
\lambda=\frac{h}{p}=\frac{h}{\sqrt{(2 m K E)}}
$$
$\lambda \propto \frac{1}{\sqrt{m}}$ as all the particles have same KE.
Since $m_e < m_p < m_{\mathrm{n}} < m_\alpha$
$$ \lambda_e > \lambda_p > \lambda_{\mathrm{n}} > \lambda_\alpha $$
$\lambda \propto \frac{1}{\sqrt{m}}$ as all the particles have same KE.
Since $m_e < m_p < m_{\mathrm{n}} < m_\alpha$
$$ \lambda_e > \lambda_p > \lambda_{\mathrm{n}} > \lambda_\alpha $$
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