The orbital angular momentum of an electron in 2s-orbital is
- $+\frac{1}{2}.\frac{h}{2 \pi}$
- zero
- $\frac{h}{2 \pi}$
- $\sqrt2.\frac{h}{2 \pi}$
The Correct Option is B
Solution and Explanation
$ \, \, \, \, \, \, \, \, L=\sqrt{l(l +1)} \frac{h}{2 \pi}=0 \, \, \, \, \, $ for 2s-electrons
$\because \, \, $ For s-orbital, l= 0.
Top Questions on momentum
- An object of mass \( m \) is projected from the origin in a vertical \( xy \)-plane at an angle \( 45^\circ \) with the x-axis with an initial velocity \( v_0 \). The magnitude and direction of the angular momentum of the object with respect to the origin, when it reaches the maximum height, will be:
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Knowing the initial position \( x_0 \) and initial momentum \( p_0 \) is enough to determine the position and momentum at any time \( t \) for a simple harmonic motion with a given angular frequency \( \omega \).
Reason (R): The amplitude and phase can be expressed in terms of \( x_0 \) and \( p_0 \).
In the light of the above statements, choose the correct answer from the options given below:- A body of mass 10 kg is moving with a speed of 5 m/s. What is the momentum of the body?
- A force acting on an object of mass 500 g changes its velocity from 200 cm/s to 0.2 m/s. The change in momentum
- A disc of mass 0.2 kg is kept floating in air without falling by vertically firing bullets each of mass 0.05 kg on the disc at the rate of 10 bullets per every second. If the bullets rebound with the same speed, then the speed of each bullet is (Acceleration due to gravity = 10 m/s$^2$)
Questions Asked in JEE Advanced exam
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity):
- JEE Advanced - 2025
- Waves and Oscillations
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
- If $$ \alpha = \int_{\frac{1}{2}}^{2} \frac{\tan^{-1} x}{2x^2 - 3x + 2} \, dx, $$ then the value of $ \sqrt{7} \tan \left( \frac{2\alpha \sqrt{7}}{\pi} \right) $ is.
(Here, the inverse trigonometric function $ \tan^{-1} x $ assumes values in $ \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) $.)- JEE Advanced - 2025
- Integral Calculus
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
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- binomial expansion formula
- The total number of real solutions of the equation $$ \theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right) $$ is
(Here, the inverse trigonometric functions $ \sin^{-1} x $ and $ \tan^{-1} x $ assume values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $(-\frac{\pi}{2}, \frac{\pi}{2})$, respectively.)- JEE Advanced - 2025
- Inverse Trigonometric Functions
Concepts Used:
Momentum
It can be defined as "mass in motion." All objects have mass; so if an object is moving, then it is called as momentum.
the momentum of an object is the product of mass of the object and the velocity of the object.
Momentum = mass • velocity
The above equation can be rewritten as
p = m • v
where m is the mass and v is the velocity.
Momentum is a vector quantity and the direction of the of the vector is the same as the direction that an object.