If the kinetic energy of a particle is increased four times, then the percentage decrease in its de Broglie wavelength will be

To solve this problem, we use the relationship between the de Broglie wavelength ($\lambda$) and the kinetic energy ($K$) of a particle:

Step 1: Formula for de Broglie wavelength

The de Broglie wavelength is given by:

λ=ph​

where $h$ is Planck's constant, and $p$ is the momentum of the particle.

The momentum can be expressed in terms of kinetic energy ($K$) as:

p=2mK

where $m$ is the mass of the particle.

Thus, the de Broglie wavelength becomes:

λ=2mK

Step 2: Analyze the change in wavelength with kinetic energy

If the kinetic energy of the particle is increased four times ($K \to 4K$), the wavelength becomes:

λ′=2m⋅4K

Step 3: Percentage decrease in wavelength

The percentage decrease in the de Broglie wavelength is:

Percentage decrease=(1−λλ′​)×100=(1−21​)×100=50%

Final Answer:

The percentage decrease in the de Broglie wavelength is 50%.

The correct option is: 50%