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If the de Broglie wavelengths associated with a proton and an $\alpha$-particle are equal, then the ratio of velocities of the proton and the $\alpha$-particle will be:
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Verified Answer
The correct answer is:
$4: 1$
$4: 1$
According to question, $\lambda_p=\lambda_\alpha$
Using, $\lambda=\frac{h}{p}=\frac{h}{m v}$
$$
\begin{aligned}
&\text { So, } \frac{h}{m_p \times v_p}=\frac{h}{m_\alpha \times v_\alpha} \\
&\Rightarrow \frac{v_p}{v_\alpha}=\frac{m_\alpha}{m_p}=\frac{4 m_p}{m_p}
\end{aligned}
$$
$(\because$ mass of $\alpha$-particle is 4 times of mass of proton)
So, $\frac{v_p}{v_\alpha}=\frac{4}{1}$; i.e., $4: 1$
Using, $\lambda=\frac{h}{p}=\frac{h}{m v}$
$$
\begin{aligned}
&\text { So, } \frac{h}{m_p \times v_p}=\frac{h}{m_\alpha \times v_\alpha} \\
&\Rightarrow \frac{v_p}{v_\alpha}=\frac{m_\alpha}{m_p}=\frac{4 m_p}{m_p}
\end{aligned}
$$
$(\because$ mass of $\alpha$-particle is 4 times of mass of proton)
So, $\frac{v_p}{v_\alpha}=\frac{4}{1}$; i.e., $4: 1$
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