Question

If the kinetic energy of a particle in motion is decreased by 36%, the increase in de Broglie wavelength of the particle is:

• A. 18%
• B. 25%
• C. 20%
• D. 32%

Hint:

When dealing with percentage changes in kinetic energy and its effect on de Broglie wavelength, remember that momentum changes as the square root of kinetic energy changes, affecting the wavelength inversely.

Answer 1

The Correct Option is B

The de Broglie wavelength \( \lambda \) is given by \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is the momentum of the particle. 
Since momentum \( p \) is related to the kinetic energy \( K \) by \( p = \sqrt{2mK} \) (where \( m \) is the mass of the particle), a decrease in \( K \) leads to a decrease in \( p \). 
Assuming kinetic energy \( K \) is decreased by 36%, then \( K' = 0.64K \). Therefore, \( p' = \sqrt{2mK'} = \sqrt{2m \cdot 0.64K} = 0.8p \). 
Since \( \lambda \propto \frac{1}{p} \), the new wavelength \( \lambda' = \frac{h}{0.8p} = 1.25\lambda \), which corresponds to a 25% increase.