Question

An α particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:

• A. √2:1
• B. 2√2:1
• C. 4√2:1
• D. 8:1

Answer 1

The Correct Option is B

To solve this problem, we need to find the ratio of linear momenta acquired by an α particle and a proton when both are accelerated from rest through the same potential difference. Here is the step-by-step explanation:

The kinetic energy (\(K.E.\)) acquired by a charged particle when accelerated through a potential difference \(V\) is given by:

1. \(K.E. = qV\)

For an α particle:

− The charge \(q_{\alpha} = 2e\) (because an α particle has two protons).

Therefore, the kinetic energy is:

1. \(K.E._{\alpha} = 2eV\)

For a proton:

− The charge \(q_{p} = e\).

Therefore, the kinetic energy is:

1. \(K.E._{p} = eV\)

From the work-energy theorem, the linear momentum \(p\) of a particle is related to its kinetic energy by:

1. \(K.E. = \frac{p^2}{2m}\)

Rearranging the equation to express the momentum, we have:

1. \(p = \sqrt{2m \cdot K.E.}\)

Substituting the kinetic energy values for both particles, we get:

− For the α particle:

1. \(p_{\alpha} = \sqrt{2m_{\alpha} \cdot 2eV}\)

− For the proton:

1. \(p_{p} = \sqrt{2m_{p} \cdot eV}\)

Now, let's calculate the ratio of their linear momenta:

1. \(\frac{p_{\alpha}}{p_{p}} = \frac{\sqrt{4m_{\alpha} eV}}{\sqrt{2m_{p} eV}}\)

Simplifying the above ratio gives us:

1. \(\frac{p_{\alpha}}{p_{p}} = \sqrt{2} \cdot \sqrt{\frac{m_{\alpha}}{m_{p}}}\)

Considering that the mass of an α particle (\(m_{\alpha}\)) is approximately 4 times the mass of a proton (\(m_{p}\)), we get:

1. \(\frac{m_{\alpha}}{m_{p}} = 4\)

Thus, the ratio simplifies to:

1. \(\frac{p_{\alpha}}{p_{p}} = \sqrt{2} \times \sqrt{4} = 2\sqrt{2}\)

The correct option is therefore \(2\sqrt{2}:1\), which is Option B.