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Step-by-Step Solution
Step 1: Write the de Broglie wavelength formula
The de Broglie wavelength $ \lambda $ of a particle is given by:
$ \lambda = \frac{h}{p} = \frac{h}{mv} $
where:
$ h $ is Planck's constant,
$ p $ is the momentum of the particle,
$ m $ is the mass of the particle,
$ v $ is the velocity of the particle.
Step 2: Express de Broglie wavelengths for proton and $ \alpha $-particle
Let $ \lambda_p $ be the de Broglie wavelength for the proton, and $ \lambda_\alpha $ be for the $ \alpha $-particle. Then:
$ \lambda_p = \frac{h}{m_p \, v_p}, \quad \lambda_\alpha = \frac{h}{m_\alpha \, v_\alpha}
Step 3: Equate the two wavelengths
Given that $ \lambda_p = \lambda_\alpha $, equate these expressions:
$ \frac{h}{m_p \, v_p} = \frac{h}{m_\alpha \, v_\alpha}
Step 4: Simplify to find the ratio of velocities
Canceling $ h $ on both sides and rearranging, we get:
$ \frac{v_p}{v_\alpha} = \frac{m_\alpha}{m_p}
We know that an $ \alpha $-particle ($ {}^4He $ nucleus) has approximately four times the mass of a proton, i.e. $ m_\alpha \approx 4 \, m_p $. Substituting,
$ \frac{v_p}{v_\alpha} = \frac{4 \, m_p}{m_p} = 4
Thus, the ratio of the velocities is:
$ v_p : v_\alpha = 4 : 1
