Question

38.5 electrons and matter waves
a non-relativistic free electron has kinetic energy k. if its wavelength doubles, its kinetic energy is

○ k.
○ k/2.
○ 2k.
○ k/4.
○ 4k.

Explanation:

Step1: Recall de Broglie wavelength and kinetic energy relation

The de Broglie wavelength \(\lambda\) of a non - relativistic particle is given by \(\lambda=\frac{h}{p}\), where \(h\) is Planck's constant and \(p\) is the momentum of the particle. For a non - relativistic electron, the kinetic energy \(K = \frac{p^{2}}{2m}\), where \(m\) is the mass of the electron. We can express \(p\) from the kinetic energy formula as \(p=\sqrt{2mK}\). Substituting this into the de Broglie wavelength formula, we get \(\lambda=\frac{h}{\sqrt{2mK}}\).

Step2: Analyze the change in wavelength

Let the initial wavelength be \(\lambda_1\) and initial kinetic energy be \(K_1 = K\). Let the final wavelength be \(\lambda_2 = 2\lambda_1\) and final kinetic energy be \(K_2\). From the wavelength formula, \(\lambda_1=\frac{h}{\sqrt{2mK_1}}\) and \(\lambda_2=\frac{h}{\sqrt{2mK_2}}\). Since \(\lambda_2 = 2\lambda_1\), we have \(\frac{h}{\sqrt{2mK_2}}=2\times\frac{h}{\sqrt{2mK_1}}\).

Step3: Solve for the final kinetic energy

We can cancel out \(h\) and \(\sqrt{2m}\) from both sides of the equation \(\frac{1}{\sqrt{K_2}}=\frac{2}{\sqrt{K_1}}\). Cross - multiplying gives \(\sqrt{K_1}=2\sqrt{K_2}\). Squaring both sides, we get \(K_1 = 4K_2\). Since \(K_1 = K\), then \(K = 4K_2\), and solving for \(K_2\) gives \(K_2=\frac{K}{4}\).

Answer:

\(K/4\) (the option corresponding to \(K/4\))