Question

De Broglie wavelength of neutrons in thermal equilibrium is (given $m_n=1.6\times {10}^{-27}kg)$

• A. $30.8/\sqrt{T}\stackrel{o}{A}$
• B. $3.08/\sqrt{T}\stackrel{o}{A}$
• C. $0.308/\sqrt{T}\stackrel{o}{A}$
• D. $0.0308/\sqrt{T}\stackrel{o}{A}$

Answer 1

The correct option is A $30.8/\sqrt{T}\stackrel{o}{A}$

The de Broglie wavelength, $\lambda =\frac{h}{\sqrt{2mkT}}$ where $h=$ Planck's constant, $m=$ mass of the particle , $k=$ Boltzmann's constant, $T=$ temperature.

So, $\lambda =\frac{6.62\times {10}^{-34}}{\sqrt{2\times (1.6\times {10}^{-27})\times (1.38\times {10}^{-23})T}}=31.5\times {10}^{-10}m/\sqrt{T}\sim 30.8/\sqrt{T}{A}^o$