Question
The de Broglie wavelength associated with an electron accelerated through a potential difference V is $\lambda_{e}$ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is $\lambda_{p}$. If their corresponding masses are $m_{e}$ and $m_{p}$, respectively, then the ratio of their de Broglie wavelengths $\left( \frac{\lambda_{e}}{\lambda_{p}} \right)$ is $\_\_\_\_$ .
Options
Solution
Sol. [Key Concept: $\lambda = \frac{h}{\sqrt{2mqV}} \propto \frac{1}{\sqrt{\ }m}$, so $\frac{\lambda_{e}}{\lambda_{p}} = \sqrt{m_{p}/m_{e}}$.]
$$\lambda = \frac{h}{\sqrt{2mqV}} \propto \frac{1}{\sqrt{\text{ }m}} $$So $\frac{\lambda_{e}}{\lambda_{p}} = \sqrt{\frac{m_{p}}{m_{e}}}$
$$\sqrt{\frac{m_{e}}{m_{p}}} $$
Create a free account to view solution
View Solution Free