If the kinetic energy of a non-relativistic electron is increased by a factor of 4, how do… — JAMB UTME

If the kinetic energy of a non-relativistic electron is increased by a factor of $4$ , how does its associated de Broglie wavelength change?

The de Broglie wavelength decreases to one-half of its original value.

The de Broglie wavelength

\lambda

of a particle is given by

\lambda = \frac{h}{p}

, where momentum

p = \sqrt{2m E_k}

. Substituting momentum into the wavelength equation yields

\lambda = \frac{h}{\sqrt{2m E_k}}

. If kinetic energy

E_k

quadruples, the denominator increases by a factor of

\sqrt{4} = 2

, which reduces the wavelength to half of its initial value.

Relate de Broglie wavelength to momentum and kinetic energy.

The de Broglie wavelength is given by

\lambda = \frac{h}{p}

. Since kinetic energy

E_k = \frac{p^2}{2m}

, momentum is

p = \sqrt{2m E_k}

. Thus,

\lambda = \frac{h}{\sqrt{2m E_k}}

Establishing the mathematical relationship between wavelength

\lambda

and kinetic energy

E_k

Apply the scaling factor of 4 to the kinetic energy.

When

E_k' = 4 E_k

, the new wavelength

\lambda'

\lambda' = \frac{h}{\sqrt{2m (4 E_k)}} = \frac{h}{2\sqrt{2m E_k}} = \frac{\lambda}{2}

Evaluating the square root factor

\sqrt{4} = 2

in the denominator.

Conclude the final ratio.

The new wavelength is half the original wavelength.

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

Relationship between de Broglie wavelength and kinetic energy

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