# NEET Questions Solved

NEET - 2017

The de-Brogile wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is

(a)$\frac{h}{\sqrt{mkT}}$

(b)$\frac{h}{\sqrt{3mkT}}$

(c)

(d)$\frac{2h}{\sqrt{mkT}}$

(b) Thinking Processs de-Brogile wavelength associated with a moving particle can be given as

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2m\left(K.E\right)}}$

At thermal equilibrium,temperature of neutron and heavy water will be same. This common temperature is given as, T. Also, we know that, kinetic energy of a particle

$KE=\frac{{p}^{2}}{2m}$

where, p=momentum of the particle

m=mass of the particle

Kinetic energy of the neutron is

$KE=\frac{3}{2}kT$

$\therefore$ de-Brogile wavelength of the neutron

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2m\left(K.E\right)}}$

$=\frac{h}{\sqrt{2m×\frac{3}{2}kT}}=\frac{h}{\sqrt{3mkT}}$

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