Question

Find the ratio $\dfrac{\lambda_\alpha}{\lambda_p}$ of the de Broglie wavelength $\lambda_\alpha$ associated with an alpha particle to the de Broglie wavelength $\lambda_p$ associated with a proton if both are moving with the same velocity.

Hint:

Use $\lambda = h/(mv)$; at equal velocity the ratio of wavelengths is the inverse ratio of masses, $m_p/m_\alpha = 1/4$.

Answer 1

Step 1: Write the de Broglie relation:
The de Broglie wavelength of a particle of mass \(m\) moving with speed \(v\) is
\[ \lambda = \frac{h}{p} = \frac{h}{mv} \]

Step 2: Apply it to both particles:
Both particles have the same velocity \(v\). For the alpha particle and the proton:
\[ \lambda_\alpha = \frac{h}{m_\alpha v}, \qquad \lambda_p = \frac{h}{m_p v} \]

Step 3: Take the ratio:
\[ \frac{\lambda_\alpha}{\lambda_p} = \frac{h/(m_\alpha v)}{h/(m_p v)} = \frac{m_p}{m_\alpha} \]
Here \(h\) and \(v\) cancel. Only the masses matter.

Step 4: Use the masses:
An alpha particle is a helium nucleus with 2 protons and 2 neutrons.
Its mass is about four times the proton mass, so \(m_\alpha \approx 4m_p\).
\[ \frac{\lambda_\alpha}{\lambda_p} = \frac{m_p}{4m_p} = \frac{1}{4} \]

Final Answer:
For the same velocity, the alpha particle has one-fourth the wavelength of the proton. \[ \boxed{\frac{\lambda_\alpha}{\lambda_p} = \frac{1}{4} \ (\text{i.e. } 1:4)} \]