De Broglie Wavelength Ratio of Proton and Alpha Particle

To find the ratio of the de Broglie wavelength of the proton to that of the $\alpha$-particle, we use the formula for the de Broglie wavelength of a particle with charge $q$ and mass $m$ accelerated from rest through a potential difference $V$:

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}$

where:

• $h$ is Planck's constant,
• $m$ is the mass of the particle,
• $q$ is the magnitude of the charge of the particle,
• $V$ is the accelerating potential difference ($V = 100\text{ V}$).

For the proton ($p$):

• Mass: $m_p \approx m$
• Charge: $q_p = e$

Thus, the de Broglie wavelength of the proton is:
$\lambda_p = \frac{h}{\sqrt{2 m_p e V}}$

For the $\alpha$-particle ($\alpha$):

• Mass: $m_\alpha \approx 4 m_p$
• Charge: $q_\alpha = 2e$

Thus, the de Broglie wavelength of the $\alpha$-particle is:
$\lambda_\alpha = \frac{h}{\sqrt{2 (4 m_p) (2e) V}} = \frac{h}{\sqrt{16 m_p e V}}$

Now, taking the ratio $\frac{\lambda_p}{\lambda_\alpha}$:

$\frac{\lambda_p}{\lambda_\alpha} = \frac{\frac{h}{\sqrt{2 m_p e V}}}{\frac{h}{\sqrt{16 m_p e V}}} = \sqrt{\frac{16 m_p e V}{2 m_p e V}} = \sqrt{8} = 2\sqrt{2}$

Calculating the value of $2\sqrt{2}$:
$\frac{\lambda_p}{\lambda_\alpha} = 2 \times 1.414 = 2.828$

Rounding to the nearest integer gives:
$\frac{\lambda_p}{\lambda_\alpha} \approx 3$