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

To find the ratio of the de Broglie wavelengths of the electron and the proton, we will use the de Broglie wavelength formula, which is given by:

\(\lambda = \frac{h}{mv}\)

where:

• \(\lambda\) is the de Broglie wavelength,
• h is the Planck's constant,
• m is the mass of the particle,
• v is the velocity of the particle.

Given that both the electron and the proton are moving with the same speed v, their wavelengths can be expressed as:

• For the electron: \(\lambda_{electron} = \frac{h}{m_e v}\)
• For the proton: \(\lambda_{proton} = \frac{h}{m_p v}\)

The ratio of their de Broglie wavelengths is:

\(\frac{\lambda_{electron}}{\lambda_{proton}} = \frac{\frac{h}{m_e v}}{\frac{h}{m_p v}} = \frac{m_p}{m_e}\)

Substituting the given mass ratio m_p = 1836 m_e, we find:

\(\frac{\lambda_{electron}}{\lambda_{proton}} = \frac{1836 m_e}{m_e} = 1836\)

Thus, the ratio of their de Broglie wavelengths \(\frac{\lambda_{electron}}{\lambda_{proton}}\) is 1836.