Question

The ratio of the velocities of ${CH}_4$ and $O_2$ molecules, so that they are associated with de-Broglie waves of equal wavelength is:

Answer 1

Correct Answer: 2

Explanation:
According to de-Broglie's equation:$\lambda =\frac{h}{mv}$Where, $\lambda =$ de-Broglie's wavelength$h=$ Planck's constant$m=$ Mass of the electron$v=$ Velocity of the electronFor ${CH}_4$, wavelength is given as:${\lambda }_{C{H}_4}=\frac{h}{m_{C{H}_4}\times v_{C{H}_4}}$For $O_2$, the wavelength is given as:${\lambda }_{O_2}=\frac{h}{m_{O_2}\times v_{O_2}}$Since, wavelength of $C{H}_4$ and $O_2$ are equal, i.e. ${\lambda }_{C{H}_4}={\lambda }_{O_2}$Or,$\frac{h}{m_{{CH}_4}\times v_{{CH}_4}}=\frac{h}{m_{O_2}\times v_{O_2}}$$\Rightarrow m_{C{H}_4}\times v_{C{H}_4}=m_{O_2}\times v_{O_2}$$\Rightarrow \frac{v_{C{H}_4}}{v{o}_2}=\frac{m_{O_2}}{m_{C{H}_4}}$Molecular mass of oxygen $=32$Molecular mass of methane $=16$$\therefore \frac{v_{C{H}_4}}{v_{O_2}}=\frac{m_{O_2}}{m_{C{H}_4}}$$=\frac{32}{16}$$=2$Hence, the correct answer is $2$.