Question

The kinetic energy of a charged particle is increased to four times of its initial value. The de Broglie wavelength associated with the particle will ______.

Options

• increase by 100% of its initial value.

• increase by 50% of its initial value.

• decrease by 25% of its initial value.

• decrease by 50% of its initial value.

Answer 1

The kinetic energy of a charged particle is increased to four times of its initial value. The de Broglie wavelength associated with the particle will decrease by 50% of its initial value.

Explanation:

The de Broglie wavelength is inversely proportional to the square root of the kinetic energy of the particle.

λ = `h/p`

= `h/(sqrt(2 m K))`

Where K is the kinetic energy.

Let initial wavelength be λ₁ = `h/(sqrt(2 m K_1))`

The kinetic energy is increased to four times, so K₂ = 4K₁.

The new wavelength λ₂ is:

λ₂ = `h/(sqrt(2m(4 K_1)))`

= `1/sqrt4(h/(sqrt(2mK_1)))`

= `lambda_1/2`    ...(i)

The change in wavelength is:

∆λ = λ₂ − λ₁

= `lambda_1/2 - lambda_1`    ...[From equation (i)]

= `-lambda_1/2`    ...(ii)

Percentage change:

`(Delta lambda)/lambda_1 xx 100`

= `(-lambda_1/2)/(lambda_1) xx 100`    ...[From equation (ii)]

= `-lambda_1/lambda_1 xx 100/2`

= −50%

The negative sign indicates a decrease.

∴ The de Broglie wavelength decreases by 50% of its initial value.