A hydrogen atom in ground state absorbs 12.09 eV of energy. The orbital angular momentum of the electron is increased by :
A hydrogen atom in ground state absorbs 12.09 eV of energy. The orbital angular momentum of the electron is increased by :
Show Hint
- 1.05 \( \times \) 10\( - \)34 Js
- 2.11 \( \times \) 10\( - \)34 Js
- 3.16 \( \times \) 10\( - \)34 Js
- 4.22 \( \times \) 10\( - \)34 Js
The Correct Option is B
Solution and Explanation
Step 2: En = -13.6/n2 = -1.51 gives n2 = 9, so n = 3.
Step 3: Angular momentum L = nh/2\(\pi\). Change \(\Delta\)L = (3 - 1)h/2\(\pi\) = 2 × 6.63 × 10-34/(2 × 3.14) = 2 × 1.055 × 10-34 = 2.11 × 10-34 J s.
Answer: (B) 2.11 × 10-34 Js
Top JEE Main Physics Questions
- 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:-
- A particle starts with an initial velocity of 10.0 m/s along x-direction and accelerates uniformly at the rate of 2.0 m/s$^2$. The time taken by the particle to reach the velocity of 60.0 m/s is
Top JEE Main Atoms Questions
- If Rydberg’s constant is \( R \), the longest wavelength of radiation in Paschen series will be \( \frac{\alpha}{7R} \), where \( \alpha = \_\_\_\_\_\_\_ \).
- The radius of the third stationary orbit of an electron in Bohr's atom is \( R \). The radius of the fourth stationary orbit will be:
- If the wavelength of the first member of the Lyman series of hydrogen is \( \lambda \). The wavelength of the second member will be:
- A hydrogen atom in ground state is given an energy of 10.2 eV. How many spectral lines will be emitted due to transition of electrons ?
- In hydrogen like system the ratio of coulombian force and gravitational force between an electron and a proton is in the order of:
Top JEE Main Questions
- 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 ______.
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,