Atomic Structure Question 258
Question: At temperature T, the average kinetic energy of any particle is $ \frac{3}{2}KT. $ The de Broglie wavelength follows the order:
Options:
A) Visible photon > Thermal neutron > Thermal electron
B) Thermal proton > Thermal electron > Visible photon
C) Thermal proton > Visible photon > Thermal electron
D) Visible photon > Thermal electron > Thermal neutron
Show Answer
Answer:
Correct Answer: D
Solution:
- Kinetic energy of any particle $ =\frac{3}{2}KT $ Also $ K.E.=\frac{1}{2}mv^{2} $
$ \frac{1}{2}mv^{2}=\frac{3}{2}KT\Rightarrow v^{2}=\frac{3KT}{m} $
$ v=\sqrt{\frac{3KT}{m}} $ De-broglie wavelength $ =\lambda =\frac{h}{mv}=\frac{h}{m\sqrt{\frac{3KT}{m}}} $
$ \lambda =\frac{h}{\sqrt{3KTm}}\lambda \propto \frac{1}{\sqrt{m}} $ Mass of electron < mass of neutron $ \lambda $ (electron) > $ \lambda $ (neutron)
BETA
