Question

the de broglie wavelength of a neutron at 27 degree celcius is lambda. what will be its wavelength at 927 degree celcius?

Answer 1

See the image below......

Answer 2

The wavelength at 927 degree Celsius $\frac{\lambda}{2}$

Solution:

At 27 degree Celsius, de Broglie wavelength = \lambda

The wavelength of de Broglie is given below:

$\lambda=\frac{h}{\sqrt{2 m E}}$

The energy of a thermal neutrons is given below:

$E = kT$

On substituting the energy on thermal neutron on wavelength of the neutron, we get,

$\lambda=\frac{h}{\sqrt{2 m k T}}$

On applying the equipartition theorem and the ideal gas equation, we render to the derivation where the lambda is equal to inverse of temperature dependent.  

Thereby,

The wavelength at 27 degree Celsius is given below:

$\frac{\lambda_{1}}{\lambda_{2}}=\frac{\frac{h}{\sqrt{2 m k T_{1}}}}{\frac{h}{\sqrt{2 m k T_{2}}}}$

Where,

λ_1= Wavelength at 27 degree Celsius.

λ_2= Wavelength at 927 degree Celsius.

$\frac{\lambda}{\lambda_{2}}=\frac{\sqrt{T_{2}}}{\sqrt{T_{1}}}$

$\frac{\lambda}{\lambda_{2}}=\frac{\sqrt{927+273}}{\sqrt{27+273}}$

$\frac{\lambda}{\lambda_{2}}=\frac{\sqrt{1200}}{\sqrt{300}}$

On calculating, we get,

$\frac{\lambda}{\lambda_{2}}=2$

$\Rightarrow \lambda_{2}=\frac{\lambda}{2}$