Question

An enclosure filled with helium is heated to a temperature of $400K$. A beam of helium emerges out of the enclosure. The mean de-broglie wavelength of this beam is :

• A. $0.44\mathring{A}$
• B. $0.63\mathring{A}$
• C. $0.77\mathring{A}$
• D. $Noneofthese$

Answer 1

The correct option is C $0.77\mathring{A}$

Given,

$T=400K$

$N_A=6.022\times {10}^{23}mo{l}^{-1}$

$k=1.38\times {10}^{-23}J/K$

$m=\frac{M}{N_A}=\frac{4\times {10}^{-3}}{6.022\times {10}^{23}}$

$m=0.664\times {10}^{-26}kg$

The mean velocity,

$V_{mean}=\sqrt{\frac{8kT}{\pi m}}$

$V_{mean}=\sqrt{\frac{8\times 1.38\times {10}^{-23}\times 400}{3.14\times 0.664\times {10}^{-26}}}$

$V_{mean}=1455.3m/s$

The mean de-broglie wavelength, $\lambda =\frac{h}{m{V}_{mean}}$

$\lambda =\frac{6.626\times {10}^{-34}}{0.664\times {10}^{-26}\times 1455.3}$

$\lambda =7\times {10}^{-11}$

$\lambda =0.77\times {10}^{-10}m$

The correct option is C.