Question

The ratio of De Broglie wavelength of molecules of $H_2$ and $He$ at ${27}^{0}C$ and ${127}^{0}C$ respectively is:

• A. $1.633$
• B. $0.612$
• C. $1.265$
• D. $0.79$

Answer 1

Answer: (A)

$1.633$

According to De-broglie,
$\lambda =\frac{h}{mu}$
$\therefore {\lambda }_{H_2}=\frac{h}{m_{H_2}\times \sqrt{\frac{3R{T}_1}{m_{H_2}}}}$
as $u_{rms}=\sqrt{\frac{3RT}{m}}$
${\lambda }_{He}=\frac{h}{m_{He}\times \sqrt{\frac{3R{T}_2}{m_{H_0}}}}$
$\therefore \frac{{\lambda }_{H_2}}{{\lambda }_{He}}=\frac{\sqrt{T_2\times m_{He}}}{\sqrt{T_1\times m_{H_2}}}$
$=\sqrt{\frac{400\times 4\times {10}^{-3}}{300\times 2\times {10}^{-3}}}=\sqrt{\frac{16}{6}}=1.633$