Question

An electron of mass $m_e$ and a proton of mass $m_p = 1836 m_e$ are moving with the same speed. The ratio of their de Broglie wavelength $\frac{\lambda_{electron}}{\lambda_{proton}}$ will be:

• A. 1
• B. 1836
• C. $\frac{1}{1836}$
• D. 918

Hint:

For particles moving at the same speed, the de Broglie wavelength is inversely proportional to their mass ($\lambda \propto 1/m$). The lighter particle will have the longer wavelength.

Answer 1

The Correct Option is B

The de Broglie wavelength ($\lambda$) of a particle is given by the formula $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is the momentum of the particle.
The momentum $p$ is given by $p = mv$, where $m$ is the mass and $v$ is the speed.
So, the de Broglie wavelength can be written as $\lambda = \frac{h}{mv}$.
For the electron, the wavelength is $\lambda_{electron} = \frac{h}{m_e v}$.
For the proton, the wavelength is $\lambda_{proton} = \frac{h}{m_p v}$.
We are given that they are moving with the same speed, $v$.
The ratio of their wavelengths is:
$\frac{\lambda_{electron}}{\lambda_{proton}} = \frac{h/(m_e v)}{h/(m_p v)} = \frac{m_p v}{m_e v} = \frac{m_p}{m_e}$.
We are given that $m_p = 1836 m_e$.
Therefore, the ratio is $\frac{\lambda_{electron}}{\lambda_{proton}} = \frac{1836 m_e}{m_e} = 1836$.