Question

A proton and an \( \alpha \)-particle are accelerated through different potentials \( V_1 \) and \( V_2 \) respectively so that they have the same de Broglie wavelengths. Find \( \frac{V_1}{V_2} \).

Hint:

For particles with the same de Broglie wavelength, the accelerating voltage is proportional to the mass of the particle.

Answer 1

The de Broglie wavelength ($\lambda$) is defined as $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is the momentum. The kinetic energy (KE) is related to momentum by $KE = \frac{p^2}{2m}$, which can be rearranged to $p = \sqrt{2mKE}$. For a charge $q$ accelerated through a potential $V$, the kinetic energy gained is $KE = qV$. If the de Broglie wavelengths of a proton ($\lambda_p$) and an alpha particle ($\lambda_{\alpha}$) are equal, then $\lambda_p = \lambda_{\alpha}$. This implies $\frac{h}{p_p} = \frac{h}{p_{\alpha}}$, so their momenta are equal: $p_p = p_{\alpha}$. Consequently, $\sqrt{2m_p KE_p} = \sqrt{2m_{\alpha} KE_{\alpha}}$. Squaring both sides yields $2m_p KE_p = 2m_{\alpha} KE_{\alpha}$, or $m_p KE_p = m_{\alpha} KE_{\alpha}$. Substituting $KE = qV$ into the equation, we get $m_p q_p V_1 = m_{\alpha} q_{\alpha} V_2$. Given that $m_{\alpha} \approx 4m_p$ and $q_{\alpha} = 2q_p$, we substitute these relationships: $m_p q_p V_1 = (4m_p)(2q_p) V_2$, which simplifies to $m_p q_p V_1 = 8m_p q_p V_2$. Dividing both sides by $m_p q_p$ gives $V_1 = 8V_2$. Therefore, the ratio of the accelerating potentials is $\frac{V_1}{V_2} = 8$.