The ratio of de-Broglie wavelengths of molecu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{m v_{rms}}$.For a gas molecule,the root mean square velocity is $v_{rms} =

Vedclass · Accepted Answer

$\sqrt{\frac{8}{3}}$

Vedclass · Answer

(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{m v_{rms}}$.For a gas molecule,the root mean square velocity is $v_{rms} = \sqrt{\frac{3kT}{m}}$.Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{3mkT}}$.Given $T_H = 27^oC = 300 K$ and $T_{He} = 127^oC = 400 K$.The masses are $m_H = 2$ units and $m_{He} = 4$ units.The ratio is $\frac{\lambda_H}{\lambda_{He}} = \sqrt{\frac{m_{He} T_{He}}{m_H T_H}} = \sqrt{\frac{4 	imes 400}{2 	imes 300}} = \sqrt{\frac{1600}{600}} = \sqrt{\frac{8}{3}}$.

The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperatures $27^oC$ and $127^oC$ respectively is

Similar Questions

The de Broglie wavelengths for an electron and a photon are $\lambda_{e}$ and $\lambda_{p}$ respectively. For the same kinetic energy $K$ of the electron and the photon,which of the following presents the correct relation between the de Broglie wavelengths of the two?

For the same objective,find the ratio of the least separation between two points to be distinguished by a microscope for light of $5000\,\mathring{A}$ and electrons accelerated through $100\,V$ used as the illuminating substance.

If the kinetic energy of an electron and a proton are equal,find the ratio of their associated de Broglie wavelengths.

The de Broglie wavelength of an electron accelerated to a potential of $ 400 \,V $ is approximately (in $\,nm$)

What does the de Broglie equation for an electron represent?

Vedclass Test Series

Exam Paper Generator

Online Exam Module