Question:
The mass of a particle is 16 times heavier than another particle and they have the same kinetic energy. The de-Broglie wavelength of the lighter particle is:
The mass of a particle is 16 times heavier than another particle and they have the same kinetic energy. The de-Broglie wavelength of the lighter particle is:
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For same kinetic energy, de-Broglie wavelength varies inversely as square root of mass.
Updated On: Jun 19, 2026
- Same as the heavier particle’s wavelength
- 2 times the heavier particle’s wavelength
- 4 times the heavier particle’s wavelength
- \( \frac{1}{4} \) times the heavier particle’s wavelength
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The Correct Option is C
Solution and Explanation
Step 1: Use de-Broglie wavelength relation.
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \]
Step 2: Same kinetic energy condition.
Given \(K\) is same for both particles, so: \[ \lambda \propto \frac{1}{\sqrt{m}} \]
Step 3: Mass relation.
Heavier particle: \[ m_H = 16 m_L \]
Step 4: Ratio of wavelengths.
\[ \frac{\lambda_L}{\lambda_H} = \sqrt{\frac{m_H}{m_L}} = \sqrt{16} = 4 \]
Step 5: Interpretation.
Lighter particle has higher wavelength.
Step 6: Final conclusion.
Thus: \[ \boxed{\lambda_L = 4 \lambda_H} \]
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \]
Step 2: Same kinetic energy condition.
Given \(K\) is same for both particles, so: \[ \lambda \propto \frac{1}{\sqrt{m}} \]
Step 3: Mass relation.
Heavier particle: \[ m_H = 16 m_L \]
Step 4: Ratio of wavelengths.
\[ \frac{\lambda_L}{\lambda_H} = \sqrt{\frac{m_H}{m_L}} = \sqrt{16} = 4 \]
Step 5: Interpretation.
Lighter particle has higher wavelength.
Step 6: Final conclusion.
Thus: \[ \boxed{\lambda_L = 4 \lambda_H} \]
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