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: