Question

Two subatomic particles 1 and 2, with the same kinetic energies, have their de-Broglie wavelengths as \( \lambda_1 \) and \( \lambda_2 \) and masses as 3 m and 6 m respectively. Determine the ratio \( \lambda_1 : \lambda_2 \).

• A. 2 : 1
• B. 1 : 2
• C. \( \sqrt{2} : 1 \)
• D. 1 : \( \sqrt{2} \)

Hint:

For particles with the same kinetic energy, the de-Broglie wavelength is inversely proportional to the square root of the particle’s mass. The heavier particle will have a shorter wavelength.

Answer 1

The Correct Option is C

Step 1: Use the de-Broglie wavelength formula.
The de-Broglie wavelength \( \lambda \) is given by: \[ \lambda = \frac{h}{p} \]
where: - \( h \) is Planck's constant,
- \( p \) is the momentum of the particle, which is related to its mass \( m \) and velocity \( v \) by: \[ p = mv. \]

Step 2: Relate the kinetic energy to the velocity.
The kinetic energy \( E_k \) of a particle is given by: \[ E_k = \frac{1}{2} mv^2. \]
Rearranging for velocity \( v \), we get: \[ v = \sqrt{\frac{2 E_k}{m}}. \]

Step 3: Compare the de-Broglie wavelengths.
The de-Broglie wavelength is then:
\[ \lambda = \frac{h}{m \sqrt{\frac{2 E_k}{m}}} = \frac{h}{\sqrt{2 m E_k}}. \]
Since both particles have the same kinetic energy, \( E_k \), the ratio of their wavelengths will depend on their masses:
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{6}{3}} = \sqrt{2}. \]

Step 4: Conclusion.
Therefore, the ratio of the de-Broglie wavelengths is \( \lambda_1 : \lambda_2 = \sqrt{2} : 1 \), and the correct answer is option (C).