The frequency of revolution of the electron in Bohr's orbit varies with n , the principal quantum number as

Year	2026	2025	2024	Total
Questions	5	8	9	22

Q53 jee_main_2024_29_january_evening Bohr's Model of Hydrogen Atom and Hydrogen Spectrum

Hydrogen atom is bombarded with electrons accelerated through a potential difference of

V

, which causes excitation of hydrogen atoms. If the experiment is being performed at

T = 0 K

, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be

(α)/(10) V

, where

α =

________.

Numerical Answer. Answer: 121 to 121

Solution

Related Formula

The energy of an electron in the $n$ $n$-th state of a hydrogen atom is:

Eₙ = -(13.6)/(n²) eV

Core Logic

At $T = 0 K$ $T = 0\text{ K}$, all hydrogen atoms are in their ground state ( $n = 1$ $n = 1$). To observe any lines in the Balmer series emission spectrum:

The emission lines must result from a transition terminating in the state $n = 2$ $n = 2$.
This requires the hydrogen electron to first be excited to at least the state $n = 3$ $n = 3$.

Thus, the bombarding electrons must have enough kinetic energy to excite the ground-state electron ( $n = 1$ $n = 1$) to the $n = 3$ $n = 3$ state.

Step 1: Calculate the Required Energy and Potential

The required energy difference is:

Δ E = E₃ - E₁ = -(13.6)/(3²) - (-(13.6)/(1²)) eV

Δ E = 13.6 (1 - (1)/(9)) eV = 13.6 \times (8)/(9) eV \approx 12.09 eV

Since the excitation energy is provided by electrons accelerated through potential $V$ $V$, the minimum potential difference required is:

V \approx 12.09 V \approx 12.1 V

Step 2: Solve for Alpha

Comparing to the given form $(α)/(10) V$ $\frac{\alpha}{10}\text{ V}$:

12.1 = (α)/(10) α = 121

Pattern Recognition

Balmer series emission requires excitation of the electron to at least $n=3$ $n=3$. Ground state $n=1$ $n=1$ to $n=3$ $n=3$ excitation energy is $12.09 eV \approx 12.1 eV$ , matching $α = 121$ $\alpha = 121$ when multiplied by $10$ $10$.

Chapter Mix

Class 12 Physics: Atoms

Q41 jee_main_2024_27_jan_morning Bohr Model of Hydrogen Atom

The radius of the third stationary orbit of an electron for a Bohr's atom is

R

. The radius of the fourth stationary orbit will be:

A. $(4)/(3)R$ $\frac{4}{3}R$
B. $(16)/(9)R$ $\frac{16}{9}R$
C. $(3)/(4)R$ $\frac{3}{4}R$
D. $(9)/(16)R$ $\frac{9}{16}R$

Solution

Related Formula

rₙ \propto (n²)/(Z)

Where $n$ $n$ is the principal quantum number and $Z$ $Z$ is the atomic number.

Core Logic

For a given atom, $Z$ $Z$ is constant, hence:

(r₄)/(r₃) = ((4)/(3))² = (16)/(9)

Given $r₃ = R$ $r_3 = R$:

Step 1: Calculate the final radius

r₄ = (16)/(9)R

Pattern Recognition

Bohr orbital dimensions scale quadratically with quantum index ( $n²$ $n^2$), rendering quick ratios straightforward via simple squaring operations.

Chapter Mix

Class 12 Physics: Atoms

Q jee_main_2024_29_jan_morning Bohr Model of Hydrogen Atom

When a hydrogen atom going from

n = 2

to

n = 1

emits a photon, its recoil speed is

(x)/(5) ~m / s

. Where

x =

_______. (Use: mass of hydrogen atom

= 1.6 \times 10⁻²⁷ ~kg

)

Numerical Answer. Answer: 17 to 17

Solution

Related Formula

By conservation of linear momentum, the momentum of the recoiling hydrogen atom matches the momentum of the emitted photon:

patom = pphoton m v = (Δ E)/(c)

Hence, the recoil speed $v$ $v$ is:

v = (Δ E)/(m c)

where $Δ E$ $\Delta E$ is the energy difference between the electronic energy levels.

Core Logic

For a hydrogen transition from $n = 2$ $n = 2$ to $n = 1$ $n = 1$:

Δ E = E₂ - E₁ = -3.4 ~eV - (-13.6 ~eV) = 10.2 ~eV

Converting energy to Joules:

Δ E = 10.2 \times 1.6 \times 10⁻¹⁹ ~J

Electronic transition energy level diagram from n=2 to n=1 for Q51

Step 1: Compute Recoil Speed

Substituting given constant values:

v = 10.2 \times 1.6 \times 10⁻¹⁹ ~J(1.6 \times 10⁻²⁷ ~kg) \times (3 \times 10⁸ ~m/s)

Step 2: Simplify Arithmetic Fractions

Cancelling out the factor $1.6$ $1.6$ from numerator and denominator:

v = 10.2 \times 10⁻¹⁹3 \times 10⁻¹⁹ = (10.2)/(3) = 3.4 ~m/s

Step 3: Solve for Variable x

Equating the computed speed to the given parameter form:

3.4 = (x)/(5) x = 3.4 \times 5 = 17

Therefore, the value of $x$ $x$ is $17$ $17$.

Pattern Recognition

Always look for clean component cancellation matches prior to raw exponential expansion. In this problem, the value $1.6$ $1.6$ in the electron charge value cancels with the structural atomic mass constant value perfectly.

Chapter Mix

Class 12 Physics: Atoms

Q37 jee_main_2024_30_january_evening Bohr Model Magnetic Moment

An electron revolving in

nth

Bohr orbit has magnetic moment

μₙ

. If

μₙ \propto nx

, the value of

x

is:

A. $2$ $2$
B. $1$ $1$
C. $3$ $3$
D. $0$ $0$

Solution

Related Formula

μₙ = i A = (e)/(T) π r²

Alternative relation using angular momentum:

μₙ = (e)/(2m) L

Core Logic

From Bohr's quantization condition, angular momentum $L$ $L$ is given by:

L = (nh)/(2π)

Therefore, the magnetic moment can be written directly as:

μₙ = (e)/(2m) ((nh)/(2π))

μₙ \propto n

Step 1: Direct Proportionality Analysis

We can also derive this via velocity and radius: $v \propto (1)/(n)$ $v \propto \frac{1}{n}$ and $r \propto n²$ $r \propto n^2$.

μₙ = (e v r)/(2) \propto ((1)/(n)) (n²) = n

Comparing with $μₙ \propto n^x$ $\mu_n \propto n^x$, we find $x = 1$ $x = 1$.

Pattern Recognition

Magnetic moment of an electron in a Bohr orbit is always directly proportional to the principal quantum number $n$ $n$. It is an integer multiple of the Bohr magneton ( $μB = (eh)/(4π m)$ $\mu_B = \frac{eh}{4\pi m}$).

Chapter Mix

Class 12 Physics: Atoms

Q36 jee_main_2024_30_jan_morning Bohr Model and Electron Energy

The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the

5th

excited state of a hydrogen atom is :

A. $4$ $4$
B. $(1)/(4)$ $\frac{1}{4}$
C. $(1)/(2)$ $\frac{1}{2}$
D. $1$ $1$

Solution

Related Formula

$KE = -E$ $\text{KE} = -E$ $PE = 2E$ $\text{PE} = 2E$

KE = (1)/(2) |PE|

Core Logic

In any allowed Bohr orbit (for any value of principal quantum number $n$ $n$), the relationship between kinetic energy (KE), potential energy (PE), and total energy (E) strictly obeys the virial theorem for a Coulombic force field.

|PE| = 2 \times KE

Step 1: Formulate the Ratio

The question asks for the ratio of the magnitude of KE to the magnitude of PE.

KE|PE| = (1)/(2)

This ratio is independent of the orbit state. Even though it is the $5th$ $5^{\text{th}}$ excited state, the ratio remains $(1)/(2)$ $\frac{1}{2}$.

Pattern Recognition

Energy relationships in Bohr orbits (and planetary motion): $K = -E = -U/2$ $K = -E = -U/2$. The state number ( $n$ $n$) is given solely as a distractor.

Chapter Mix

Class 12 Physics: Atoms

The frequency of revolution of the electron in Bohr's orbit varies with n$n$ , the principal quantum number as

Solution & Explanation

Related Formula

Core Logic

Step 1: Verification

Pattern Recognition

Chapter Mix

Reference Study Guides

More Atoms Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Calculate the Required Energy and Potential

Step 2: Solve for Alpha

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the final radius

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Recoil Speed

Step 2: Simplify Arithmetic Fractions

Step 3: Solve for Variable x

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Direct Proportionality Analysis

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Formulate the Ratio

Pattern Recognition

Chapter Mix

More Atoms Questions — jee_main_2025_28_jan_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

The frequency of revolution of the electron in Bohr's orbit varies with $n$ $n$ , the principal quantum number as