Question 147: Dual Nature of Matter and Radiation

Question

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 :

Accepted Answer

$$
\lambda=\frac{h}{p}=\frac{h}{\sqrt{(2 m K E)}}
$$
$\lambda \propto \frac{1}{\sqrt{m}}$ as all the particles have same KE.
Since $m_e  \lambda_p > \lambda_{\mathrm{n}} > \lambda_\alpha
$$

Answer

$$\lambda$$p = $$\lambda$$n > $$\lambda$$e > $$\lambda$$a

Answer

$$\lambda$$a < $$\lambda$$n < $$\lambda$$p < $$\lambda$$e

Answer

$$\lambda$$e < $$\lambda$$p = $$\lambda$$n > $$\lambda$$a

Answer

$$\lambda$$e = $$\lambda$$p = $$\lambda$$n = $$\lambda$$a

Dual Nature of Matter and Radiation

A proton, a neutron, an electron and an $\alpha particle have same energy. If \lambdap, \lambdan, \lambdae and \lambdaa are the de Broglie's wavelengths of proton, neutron, electron and \alpha$ particle respectively, then choose the correct relation from the following :

Choose the correct answer: