Question

38.5 electrons and matter waves
a free electron and a free proton have the same non - relativistic kinetic energy. this means that, compared to the matter wave associated with the proton, the matter wave associated with the electron has

• impossible to compare as the particles are different.
• a longer wavelength.
• a shorter wavelength.
• the same wavelength.

Explanation:

1. Recall the de Broglie wavelength formula: $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is momentum.
2. For non - relativistic motion, kinetic energy $K=\frac{p^{2}}{2m}$, so $p = \sqrt{2mK}$.
3. Substitute $p$ into the de Broglie formula: $\lambda=\frac{h}{\sqrt{2mK}}$.
4. Given that the electron ($m_e$) and proton ($m_p$) have the same kinetic energy ($K$), and $m_e
5. From $\lambda=\frac{h}{\sqrt{2mK}}$, when $K$ is constant, a smaller mass $m$ leads to a larger $\lambda$ (since $m$ is in the denominator under the square root). Since the mass of the electron is less than the mass of the proton, the wavelength of the matter wave associated with the electron is longer than that associated with the proton.

Answer:

B. a longer wavelength.