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:
- √2:1
- 2√2:1
- 4√2:1
- 8:1
The Correct Option is B
Approach Solution - 1
To solve this problem, we need to determine the ratio of the linear momenta acquired by an α particle and a proton when both are accelerated from rest through the same potential difference. Let's break down the steps:
Step-by-step Solution:
- When a charged particle is accelerated through a potential difference \( V \), the work done on the particle is converted into kinetic energy. The kinetic energy \( K \) gained by the particle is given by: \(K = qV\), where \( q \) is the charge of the particle.
- The expression for kinetic energy is also given by: \(K = \frac{1}{2}mv^2\), where \( m \) is the mass and \( v \) is the velocity of the particle.
- From the above two equations, we get: \(qV = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2qV}{m}}\)
- The linear momentum \( p \) of the particle is: \(p = mv = m\sqrt{\frac{2qV}{m}} = \sqrt{2mqV}\)
- Now, compute the momenta for both the α particle and the proton:
- For the α particle (He\(^{2+}\)), the charge \( q \) is 2e, and the mass \( m \) is approximately 4 times that of the proton. Therefore, the momentum for the α particle is: \(p_{\alpha} = \sqrt{2 \cdot 4m_p \cdot 2e \cdot V} = \sqrt{16m_peV}\)
- For the proton, the charge \( q \) is e and the mass \( m \) is \( m_p \). Therefore, the momentum for the proton is: \(p_p = \sqrt{2m_peV}\)
- Ratio of the linear momenta of α particle to proton: \(\frac{p_{\alpha}}{p_p} = \frac{\sqrt{16m_peV}}{\sqrt{2m_peV}} = \frac{\sqrt{16}}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\)
Therefore, the ratio of linear momenta acquired by the α particle and the proton is 2√2:1.
Approach Solution -2
The ratio of linear momenta acquired by above two particles,
\(\frac{pα}{pp}=\frac{\sqrt{2(4m)(2eV)}}{{\sqrt{2(m)(eV)}}}\)
\(=\frac{\sqrt{16}}{√2}\)
=\(\frac{4}{√2}\)
\(=\frac{2√2}{1}\)
So, the correct option is (B): \(\frac{2√2}{1}\)
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main centripetal acceleration Questions
- If the ratio of centripetal acceleration of two particles moving on the same circular path is \(3: 4\). Find the ratio of their speed.
A body starts moving from rest with constant acceleration and covers displacement \(S_1\) in the first \((p - 1)\) seconds and \(S_2\) in the first \(p\) seconds. The displacement \(S_1 + S_2\) will be made in time:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
Concepts Used:
Types of Differential Equations
There are various types of Differential Equation, such as:
Ordinary Differential Equations:
Ordinary Differential Equations is an equation that indicates the relation of having one independent variable x, and one dependent variable y, along with some of its other derivatives.
\(F(\frac{dy}{dt},y,t) = 0\)
Partial Differential Equations:
A partial differential equation is a type, in which the equation carries many unknown variables with their partial derivatives.
Linear Differential Equations:
It is the linear polynomial equation in which derivatives of different variables exist. Linear Partial Differential Equation derivatives are partial and function is dependent on the variable.
Homogeneous Differential Equations:
When the degree of f(x,y) and g(x,y) is the same, it is known to be a homogeneous differential equation.
\(\frac{dy}{dx} = \frac{a_1x + b_1y + c_1}{a_2x + b_2y + c_2}\)
Read More: Differential Equations