Quantum Physics: Question 9

Worked solution

Part (a): Show that the kinetic energy is 8.00 × 10⁻¹⁷ J

A particle of charge magnitude $e$ accelerated from rest through a potential difference $U$ gains kinetic energy equal to the work done on it by the electric field: $E_k = eU$

This is the same for the electron and the proton, since both have charge of magnitude $e=1.60\times10^{-19}\text{ C}$ and are accelerated through the same $U=500\text{ V}$: $E_k = 1.60\times10^{-19}\times500 = 8.00\times10^{-17}\text{ J}$

Recompute as a check: $8.00\times10^{-17}/500=1.60\times10^{-19}\text{ C}$, matching $e$. So $E_k=8.00\times10^{-17}\text{ J}$, as required to show.

Part (b): Momentum of the electron

Since $E_k=\tfrac12mv^2=\dfrac{p^2}{2m}$, the momentum is: $p_e=\sqrt{2m_eE_k}$

Substituting $m_e=9.11\times10^{-31}\text{ kg}$ and $E_k=8.00\times10^{-17}\text{ J}$: $p_e=\sqrt{2\times9.11\times10^{-31}\times8.00\times10^{-17}}=\sqrt{1.458\times10^{-46}}=1.21\times10^{-23}\text{ kg m s}^{-1}$

Recompute as a check: $(1.21\times10^{-23})^2=1.46\times10^{-46}$, matching the value under the square root above. Consistent.

Part (c): Momentum of the proton

Using the same relation with the proton mass $m_p=1.67\times10^{-27}\text{ kg}$: $p_p=\sqrt{2m_pE_k}=\sqrt{2\times1.67\times10^{-27}\times8.00\times10^{-17}}=\sqrt{2.672\times10^{-43}}=5.17\times10^{-22}\text{ kg m s}^{-1}$

Recompute as a check: $(5.17\times10^{-22})^2=2.67\times10^{-43}$, matching the value under the square root above. Consistent. Note that $p_p$ is roughly $43$ times larger than $p_e$, since $p\propto\sqrt{m}$ for fixed $E_k$ and $\sqrt{m_p/m_e}=\sqrt{1.67\times10^{-27}/9.11\times10^{-31}}\approx43$.

Part (d): De Broglie wavelengths

For the electron: $\lambda_e=\frac{h}{p_e}=\frac{6.63\times10^{-34}}{1.21\times10^{-23}}=5.49\times10^{-11}\text{ m}$

For the proton: $\lambda_p=\frac{h}{p_p}=\frac{6.63\times10^{-34}}{5.17\times10^{-22}}=1.28\times10^{-12}\text{ m}$

Recompute as a check: $\lambda_e/\lambda_p=(5.49\times10^{-11})/(1.28\times10^{-12})\approx42.9$, matching the momentum ratio $p_p/p_e\approx43$ found in part (c) (since $\lambda\propto1/p$). Consistent.

Part (e): Why is the proton’s wavelength smaller?

Both particles have the same kinetic energy $E_k$, but the proton is roughly $1800$ times more massive than the electron. From $p=\sqrt{2mE_k}$, a larger mass gives a larger momentum for the same $E_k$, so the proton has a much larger momentum than the electron. Since the de Broglie wavelength is $\lambda=h/p$, a larger momentum corresponds to a shorter wavelength. This is why the (more massive, higher-momentum) proton has the smaller de Broglie wavelength, even though its kinetic energy is the same as the electron’s.

Final answers

• (a) $E_k=\boxed{8.00\times10^{-17}}\text{ J}$
• (b) $p_e=\boxed{1.21\times10^{-23}}\text{ kg m s}^{-1}$
• (c) $p_p=\boxed{5.17\times10^{-22}}\text{ kg m s}^{-1}$
• (d) $\lambda_e=\boxed{5.49\times10^{-11}}\text{ m}$, $\lambda_p=\boxed{1.28\times10^{-12}}\text{ m}$
• (e) The proton’s larger mass gives it a larger momentum for the same $E_k$, and since $\lambda=h/p$, this larger momentum gives it the smaller de Broglie wavelength.