© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write down the relevant formula for momentum
When a charged particle (charge $q$, mass $m$) is accelerated by a potential difference $V$, the kinetic energy acquired by the particle is given by:
$ qV = \frac{p^2}{2m}.$
From this, the momentum $p$ can be found as:
$ p = \sqrt{2 m q V}.$
Step 2: Express de Broglie wavelength for each particle
The de Broglie wavelength $\lambda$ of a particle is given by:
$ \lambda = \frac{h}{p},$
where $h$ is Planck's constant. Substituting $p = \sqrt{2 m q V}$, we get:
$ \lambda = \frac{h}{\sqrt{2 m q V}}.$
Step 3: Determine the ratio of wavelengths
Let $\lambda_\alpha$ be the wavelength of the $\alpha$ particle and $\lambda_p$ be the wavelength of the proton. Their ratio is:
$ \frac{\lambda_p}{\lambda_\alpha}
= \frac{\frac{h}{\sqrt{2 m_p q_p V}}}{\frac{h}{\sqrt{2 m_\alpha q_\alpha V}}}
= \frac{\sqrt{2 m_\alpha q_\alpha V}}{\sqrt{2 m_p q_p V}}
= \sqrt{\frac{m_\alpha \,q_\alpha}{m_p \,q_p}}.$
Step 4: Substitute values for the alpha particle and proton
The mass of an alpha particle is $m_\alpha = 4m_p$ (approximately 4 times that of the proton).
The charge of an alpha particle is $q_\alpha = 2e$ (twice the elementary charge $e$).
The proton has mass $m_p$ and charge $q_p = e$.
Therefore,
$ \frac{\lambda_p}{\lambda_\alpha}
= \sqrt{\frac{(4m_p)\,(2e)}{m_p\,e}}
= \sqrt{8}
= 2\sqrt{2}.$
Step 5: Numerical value and conclusion
The approximate numerical value of $2\sqrt{2}$ is about $2.828$, which rounds to $2.8$. Thus, the ratio
$ \frac{\lambda_p}{\lambda_\alpha} \approx 2.8.$
Answer: 2.8
