Question

De Broglie wavelength of a neutron at $927{}^\circ C$ is $\lambda$. What will be its wavelength at $27{}^\circ C$?
(A) $\dfrac{\lambda }{2}$
(B) $\lambda$
(C) $2\lambda$
(D) $4\lambda$

Answer 1

Try to recall that the kinetic energy of a particle is directly proportional to its absolute temperature and De-Broglie wavelength of a particle is inversely proportional to its kinetic energy. Now, by using this you can easily find the correct option from the given ones.

Complete step by step solution:
- We know that De Broglie hypothesis says that all matter has both particle and wave nature. This was called a hypothesis because there was no evidence for it when it was proposed, only analogies with existing theories.
-According to De-Broglie equation, Eq. 1:
$\lambda =\dfrac{h}{mv}$
Where $\lambda$ is the wavelength, $h$ is Planck’s constant, $m$ is the mass of the particle (here, neutron) which is moving at velocity $v$.
-We know that kinetic energy of a particle is defined as, Eq. 2:
$K=\dfrac{1}{2}m{{v}^{2}}$
-Also, kinetic energy, Eq. 3:
$K=\dfrac{3}{2}kT$
where $k$ is Boltzmann’s constant and $T$ is the absolute temperature of the particle.
Equating equation 2 and 3 we get,
$v=\sqrt{\dfrac{3kT}{m}}$
Now, putting the value of v in equation 1 we get, Eq. 4:
$\lambda =\dfrac{h}{\sqrt{3mkT}}$
Coming to the question when $T=927+273=1200K$
$\lambda =\dfrac{h}{\sqrt{3mk\times 1200}}$
Let De-Broglie wavelength be $x$ when $T=27+273=300K$, we get, Eq. 5:
$x=\dfrac{h}{\sqrt{3mk\times 300}}$
Dividing equation 4 by 5

& \dfrac{\lambda }{x}=\sqrt{\dfrac{300}{1200}} \\\ & \Rightarrow \dfrac{\lambda }{x}=\dfrac{1}{2} \\\ & x=2\lambda \\\ \end{aligned}$$ **Therefore, from above we can conclude that option C is the correct option to the given question.** **Note:** Note that the significance of de Broglie equation lies in the fact that it relates the particle character with the wave character of matter. Also, you should remember that although de Broglie equation is applicable to all material objects it has significance only in the case of microscopic particles.