Question

38.5 electrons and matter waves
a free electron and a free proton have the same non - relativistic speed. 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.
• the same wavelength.
• a shorter wavelength.
• a longer wavelength.

Explanation:

Step1: Recall de Broglie wavelength formula

The de Broglie wavelength formula is $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p = mv$ is the momentum (mass times velocity) for non - relativistic motion.

Step2: Analyze the masses of electron and proton

The mass of an electron ($m_e$) is much smaller than the mass of a proton ($m_p$), i.e., $m_e\ll m_p$. Given that the speed $v$ of the electron and proton is the same (non - relativistic), the momentum $p = mv$. So, $p_e=m_ev$ and $p_p = m_pv$. Since $m_e < m_p$, we have $p_e < p_p$.

Step3: Relate momentum to wavelength

From $\lambda=\frac{h}{p}$, since $h$ is a constant, and $p_e < p_p$, we can say that $\lambda_e=\frac{h}{p_e}$ and $\lambda_p=\frac{h}{p_p}$. If $p_e < p_p$, then $\frac{h}{p_e}>\frac{h}{p_p}$, which means $\lambda_e>\lambda_p$. So the matter wave associated with the electron has a longer wavelength than that associated with the proton.

Answer:

a longer wavelength.