If the kinetic energy of a non-relativistic free electron doubles, the frequency of its wave function changes by the factor:
A)1/
B)1/2
C)1/4

Question

If the kinetic energy of a non-relativistic free electron doubles, the frequency of its wave function changes by the factor:
A)1/ 
B)1/2 
C)1/4

Quizplus · Accepted Answer

According to the de Broglie wavelength equation, $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$, where $h$ is Planck's constant, $p$ is the momentum, $m$ is the mass, and $E$ is the kinetic energy. Since both $m$ and $h$ are constants, we can simplify the equation to $\lambda \propto \frac{1}{\sqrt{E}}$. Doubling the kinetic energy results in halving the wavelength (since $\lambda \propto \frac{1}{\sqrt{E}}$), which corresponds to doubling the frequency (since $f = \frac{c}{\lambda}$, where $c$ is the speed of light). Therefore, the answer is E, 2.

If the Kinetic Energy of a Non-Relativistic Free Electron Doubles