Question

A particle is moving with a velocity four times the velocity of an electron. If the de Broglie wavelength of the particle is \(1.5 \times 10^{-4}\) times the de Broglie wavelength of the electron, then the mass of the particle is (mass of electron \(= 9 \times 10^{-31} \, {kg}\)):

• A. \(1.5 \times 10^{-31} \, {kg}\)
• B. \(1.5 \times 10^{-27} \, {kg}\)
• C. \(2.25 \times 10^{-27} \, {kg}\)
• D. \(2.85 \times 10^{-31} \, {kg}\)

Hint:

Use \(\lambda = \frac{h}{mv}\) and relate the ratios of wavelengths and velocities to find unknown mass.

Answer 1

The Correct Option is B

The de Broglie wavelength \(\lambda\) of a particle is given by:
\[ \lambda = \frac{h}{m v} \]

Let
- \(\lambda_e\) = de Broglie wavelength of electron,
- \(\lambda_p\) = de Broglie wavelength of particle,
- \(m_e = 9 \times 10^{-31} \, \text{kg}\) = mass of electron,
- \(m_p\) = mass of particle,
- \(v_e\) = velocity of electron,
- \(v_p = 4 v_e\).

Given:
\[ \lambda_p = 1.5 \times 10^{-4} \times \lambda_e \]

Using the formula for wavelength:
\[ \lambda_p = \frac{h}{m_p v_p}, \quad \lambda_e = \frac{h}{m_e v_e} \]

Dividing these two:
\[ \frac{\lambda_p}{\lambda_e} = \frac{m_e v_e}{m_p v_p} \]

Substitute values:
\[ 1.5 \times 10^{-4} = \frac{m_e v_e}{m_p \times 4 v_e} = \frac{m_e}{4 m_p} \]

Rearranged:
\[ m_p = \frac{m_e}{4 \times 1.5 \times 10^{-4}} = \frac{9 \times 10^{-31}}{6 \times 10^{-4}} = 1.5 \times 10^{-27} \, \text{kg} \]