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Chapter 17: Problem 2428

A proton, a deuteron and an \(\propto\) -particle having the same momentum, enters a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is (A) \(1: 2: 8\) (B) \(1: 2: 4\) (C) \(1: 1: 2\) (D) None of these

Short Answer

Expert verified
The ratio of deflections suffered by a proton, deuteron, and an \(\alpha\)-particle having the same momentum when entering a region of uniform electric field between the parallel plates of a capacitor is \(1 : 2 : 8\).

Step by step solution

01

Define the given quantities

In this problem, we are given that the momentum is the same for all the particles (proton, deuteron, and \(\alpha\)-particle). Let the momentum be P.
02

Find the force acting on each particle

Let's denote the electric field between the parallel plates of a capacitor as E, and the charge of a proton as e. Then the force acting on each particle can be expressed as: - Proton: \(F_p = eE\). - Deuteron: \(F_d = eE\), electric charge of deuteron is also e as it's a nucleus with one proton and one neutron. - \(\alpha\)-particle: \(F_\alpha = 2eE\), electric charge of an \(\alpha\)-particle (Helium nucleus) is 2e because it consists of 2 protons and 2 neutrons.
03

Calculate acceleration of each particle

Now, we will find the acceleration for each particle by applying Newton's second law. For each particle: - Proton: \(a_p = F_p/m_p = eE/m_p\). - Deuteron: \(a_d = F_d/(2 m_p) = (eE)/(2 m_p)\). - \(\alpha\)-particle: \(a_\alpha = F_\alpha /(4 m_p) = (2eE)/(4 m_p) = (eE)/(2 m_p)\).
04

Find the time to cross the uniform electric field region

Since all three particles have the same momentum, their velocities perpendicular to the electric field will be the same. Let the distance between the parallel plates be d and the time to cross the electric field region be t. Then, we can write the following equation for each particle: - Proton: \(t_p = m_p d / P\). - Deuteron: \(t_d = 2m_p d / P\). - \(\alpha\)-particle: \(t_\alpha = 4m_p d / P\). Since the problem asks for the ratio of their deflections, we will consider the time taken by the proton as a reference. Then, the time ratios are given by: \(t_p : t_d : t_\alpha = 1 : 2 : 4\).
05

Calculate the deflections of each particle

The deflection of each particle can be calculated using the formula: \(\text{Deflection} = \frac{1}{2}a_t t^2\), where a_t is the acceleration of the particle and t is the time taken for the particle to cross the electric field region. - Proton: \(\text{Deflection}_p = \frac{1}{2}(eE/m_p) \cdot (t_p)^2\). - Deuteron: \(\text{Deflection}_d = \frac{1}{2}(eE/(2m_p)) \cdot (2 t_p)^2\). - \(\alpha\)-particle: \(\text{Deflection}_\alpha = \frac{1}{2}(eE/(2m_p)) \cdot (4 t_p)^2\).
06

Determine the ratio of the deflections

Dividing each deflection by the proton's deflection, we get the ratio as: \(\frac{\text{Deflection}_p}{\text{Deflection}_p} : \frac{\text{Deflection}_d}{\text{Deflection}_p} : \frac{\text{Deflection}_\alpha}{\text{Deflection}_p} = 1 : 2 : 8\). Hence, the ratio of deflections of proton, deuteron, and \(\alpha\)-particle is \(1 : 2 : 8\), which corresponds to option (A).

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Most popular questions from this chapter

If \(\propto\) -particle and proton are accelerated through the same potential difference, then the ratio of de Brogile wavelength of \(\propto\) -particle and proton is \(\ldots \ldots \ldots\) (A) \((1 / \sqrt{2})\) (B) \(\sqrt{2}\) (C) \(\\{1 /(2 \sqrt{2})\\}\) (D) \(2 \sqrt{2}\)

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