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Step-by-Step Solution
Step 1: Recall the de Broglie Relation
The de Broglie wavelength of a particle is given by:
$ \lambda = \frac{h}{p}
$
where $ \lambda $ is the wavelength, $ h $ is Planck’s constant, and $ p $ is the momentum of the particle.
Step 2: Express Momentum in Terms of Mass and Velocity
The momentum $ p $ of a particle is:
$ p = m \, v
$
Hence, the de Broglie wavelength becomes:
$ \lambda = \frac{h}{m \, v}
$
Step 3: Assign Symbols to Each Particle
Let the subscript “p” represent the proton, and “α” represent the α-particle. According to the given data:
For the proton:
$ \lambda_p = \frac{h}{m_p\,v_p} = \lambda $
For the α-particle:
$ \lambda_\alpha = \frac{h}{m_\alpha\,v_\alpha} = 2\lambda $
Step 4: Form the Ratio of the Wavelength Equations
We set up the ratio:
$ \frac{\lambda_\alpha}{\lambda_p}
=
\frac{\frac{h}{m_\alpha \, v_\alpha}}{\frac{h}{m_p \, v_p}}
=
\frac{2 \lambda}{\lambda}
=
2
$
Simplifying this ratio gives:
$ \frac{m_p \, v_p}{m_\alpha \, v_\alpha}
= 2
\quad \Longrightarrow \quad
\frac{v_p}{v_\alpha}
= \frac{2 \, m_\alpha}{m_p}
Step 5: Use the Mass Relation Between Proton and α-Particle
Typically, $ \alpha$-particle mass is about four times the mass of a proton:
$ m_\alpha \approx 4 \, m_p
$
Substituting this into the velocity ratio:
$ \frac{v_p}{v_\alpha}
=
\frac{2 \times 4 \, m_p}{m_p}
=
8
Step 6: State the Final Velocity Ratio
The ratio of the proton’s velocity to that of the α-particle is:
$ v_p : v_\alpha = 8 : 1
This confirms the correct answer.
