The de Broglie wavelength of a neutron when its kinetic energy is $K$ , is $λ$ . What will be its wavelength when its kinetic energy is $4 K$ ?

$\frac{λ}{4}$
$\frac{λ}{2}$
$2 λ$
$4 λ$

Question

The de Broglie wavelength of a neutron when its kinetic energy is K- is lambda- What will be its wavelength when its kinetic energy is 4K - dfrac-lambda-4-dfrac-lambda-2-2lambda4lambda

Toppr Admin · Accepted Answer

Kinetic energy of neutron-12mv2-K-x27F9-v-x221A-2KmDe-broglie wavelength of neturon-hp-hmv-h-x221A-2mK-x221D-1-x221A-KThus when kinetic energy becomes 4K- the de-broglie wavelength becomes -x3BB-2-

The de Broglie wavelength of a neutron when its kinetic energy is K, is λ. What will be its wavelength when its kinetic energy is 4K? λ4λ22λ4λ

Kinetic energy of neutron=12mv2=K⟹v=√2KmDe-broglie wavelength of neturon=hp=hmv=h√2mK∝1√KThus when kinetic energy becomes 4K, the de-broglie wavelength becomes λ2.

The de Broglie wavelength of a neutron when its kinetic energy is $K$ , is $λ$ . What will be its wavelength when its kinetic energy is $4 K$ ?

$\frac{λ}{4}$
$\frac{λ}{2}$
$2 λ$
$4 λ$

Kinetic energy of neutron= $\frac{1}{2} m v^{2} = K$
$⟹ v = \sqrt{\frac{2 K}{m}}$
De-broglie wavelength of neturon= $\frac{h}{p}$
$= \frac{h}{m v} = \frac{h}{\sqrt{2 m K}} \propto \frac{1}{\sqrt{K}}$
Thus when kinetic energy becomes $4 K$ , the de-broglie wavelength becomes $\frac{λ}{2}$ .