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

De Broglie Wavelength Relation:
The de Broglie wavelength is given by:
$\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is the momentum. The kinetic energy (KE) is given by:
\[ KE = \frac{p^2}{2m} \implies p = \sqrt{2mKE} \] The kinetic energy gained by a charge $q$ accelerated through a potential $V$ is:
\[ KE = qV \] Since the de Broglie wavelengths are equal:
$\lambda_p = \lambda_{\alpha}$. Therefore, $\frac{h}{p_p} = \frac{h}{p_{\alpha}}$, which implies $p_p = p_{\alpha}$. \[ \sqrt{2m_p KE_p} = \sqrt{2m_{\alpha} KE_{\alpha}} \] Squaring both sides gives:
\[ 2m_p KE_p = 2m_{\alpha} KE_{\alpha} \] \[ m_p KE_p = m_{\alpha} KE_{\alpha} \] Substituting $KE = qV$: \[ m_p q_p V_1 = m_{\alpha} q_{\alpha} V_2 \] We know that the mass of the alpha particle is approximately four times the mass of the proton, and the charge of the alpha particle is twice the charge of the proton:
$m_{\alpha} \approx 4m_p$ and $q_{\alpha} = 2q_p$. Substituting these values:
\[ m_p q_p V_1 = (4m_p)(2q_p) V_2 \] \[ m_p q_p V_1 = 8m_p q_p V_2 \] Dividing both sides by $m_p q_p$: \[ V_1 = 8V_2 \] Therefore: \[ \frac{V_1}{V_2} = 8 \]