Question 126: 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 $\alpha particle have same energy. If \lambda p , \lambda n , \lambda e and \lambda a are the de Broglie's wavelengths of proton, neutron, electron and \alpha$ particle respectively, then choose the correct relation from the following :

Choose the correct answer: