If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and (1/4pi 0) = 9 x 109 in SI units)

By energy conservation, the entire kinetic energy of the alpha particle gets converted to electrostatic potential energy at the distance of closest approach (r0). Ki + Ui = Kf + Uf Initial kinetic energy Ki = 7.7 MeV = 7.7 x 106 x 1.6 x 10-19 J. Distance of closest approach problem: Simply equate initial Kinetic Energy (in Joules) to Potential Energy k(Z1e)(Z2e)/r0. Alpha particle has charge 2e. Class 12 Physics: Atoms

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and (1/4pi 0) = 9 x 109 in SI units)

By energy conservation, the entire kinetic energy of the alpha particle gets converted to electrostatic potential energy at the distance of closest approach (r0). Ki + Ui = Kf + Uf Initial kinetic energy Ki = 7.7 MeV = 7.7 x 106 x 1.6 x 10-19 J. Distance of closest approach problem: Simply equate initial Kinetic Energy (in Joules) to Potential Energy k(Z1e)(Z2e)/r0. Alpha particle has charge 2e. Class 12 Physics: Atoms

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and (1/4pi 0) = 9 x 109 in SI units)

By energy conservation, the entire kinetic energy of the alpha particle gets converted to electrostatic potential energy at the distance of closest approach (r0). Ki + Ui = Kf + Uf Initial kinetic energy Ki = 7.7 MeV = 7.7 x 106 x 1.6 x 10-19 J. Distance of closest approach problem: Simply equate initial Kinetic Energy (in Joules) to Potential Energy k(Z1e)(Z2e)/r0. Alpha particle has charge 2e. Class 12 Physics: Atoms

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and (1/4pi 0) = 9 x 109 in SI units)

By energy conservation, the entire kinetic energy of the alpha particle gets converted to electrostatic potential energy at the distance of closest approach (r0). Ki + Ui = Kf + Uf Initial kinetic energy Ki = 7.7 MeV = 7.7 x 106 x 1.6 x 10-19 J. Distance of closest approach problem: Simply equate initial Kinetic Energy (in Joules) to Potential Energy k(Z1e)(Z2e)/r0. Alpha particle has charge 2e. Class 12 Physics: Atoms

Solution & Explanation

### Related Formula

K_textinitial = U_textclosest approach

K = frac14pi epsilon_0 frac(2e)(Ze)r_0

### Core Logic By energy conservation, the entire kinetic energy of the alpha particle gets converted to electrostatic potential energy at the distance of closest approach (

r_0

K_i + U_i = K_f + U_f

K_i + 0 = 0 + frac14piepsilon_0 frac(2e)(79e)r_0

### Step 1: Convert Energy and Solve Initial kinetic energy

K_i = 7.7text MeV = 7.7 times 10^6 times 1.6 times 10^-19text J

7.7 times 10^6 times 1.6 times 10^-19 = frac9 times 10^9 times (2 times 1.6 times 10^-19) times (79 times 1.6 times 10^-19)r_0

r_0 = frac9 times 10^9 times 2 times 79 times (1.6 times 10^-19)^27.7 times 10^6 times 1.6 times 10^-19

r_0 = frac9 times 10^9 times 158 times 1.6 times 10^-197.7 times 10^6

r_0 = frac2275.2 times 10^-107.7 times 10^6 = 295.48 times 10^-16text m approx 2.95 times 10^-14text m

### Pattern Recognition Distance of closest approach problem: Simply equate initial Kinetic Energy (in Joules) to Potential Energy

k(Z_1e)(Z_2e)/r_0

. Alpha particle has charge

2e

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

More Atoms Previous-Year Questions — Page 2

Q16 jee_main_2025_07_april_morning Hydrogen Spectrum

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is.

A. $5:36$ $5:36$
B. $5:27$ $5:27$
C. $3 : 4$ $3 : 4$
D. $27:5$ $27:5$

Solution

### Related Formula The wavelength

lambda

for a transition in a hydrogen atom is given by the Rydberg formula:

frac1lambda = R left( frac1n_1^2 - frac1n_2^2 right)

To find the largest wavelength (minimum energy transition), select the adjacent higher shell

n_2 = n_1 + 1

. ### Core Logic - **Lyman Series largest wavelength (

lambda_L

)**: Transition from

n = 2 to 1

frac1lambda_L = R left( frac11^2 - frac12^2 right) = frac3R4 implies lambda_L = frac43R

- **Balmer Series largest wavelength (

lambda_B

)**: Transition from

n = 3 to 2

frac1lambda_B = R left( frac12^2 - frac13^2 right) = R left( frac14 - frac19 right) = frac5R36 implies lambda_B = frac365R

### Step 1: Ratio Calculation Now, compute the ratio of the wavelengths:

fraclambda_Llambda_B = fracfrac43Rfrac365R = frac43 times frac536 = frac527

### Pattern Recognition Sees: Ratio of largest wavelengths of series. Shortcut: The largest wavelength in a series starting at ground level

n_1

is

\lambda \propto \frac{n_1^2 (n_1+1)^2}{2n_1 + 1}

. For Lyman (

n_1=1

):

\lambda_L \propto \frac{4}{3}

. For Balmer (

n_1=2

):

\lambda_B \propto \frac{36}{5}

. Ratio:

\frac{4/3}{36/5} = \frac{5}{27}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

Q20 jee_main_2025_29_jan_evening Hydrogen Spectrum

The number of spectral lines emitted by atomic hydrogen that is in the

4^textth

energy level, is:

A. $6$ $6$
B. $0$ $0$
C. $3$ $3$
D. $1$ $1$

Solution

### Related Formula

N = fracn(n - 1)2

where

n

is the principal quantum number of the starting energy level. ### Core Logic For a hydrogen sample initially in the

n = 4

level, the possible downward transition pathways to reach the ground state (

n=1

) are:

Hydrogen Spectrum Transitions diagram for Q20 - JEE Main 2025 Evening

Using the combination formula for all transitions:

N = frac4(4 - 1)2 = frac4 times 32 = 6

Thus, 6 distinct spectral lines are generated. ### Pattern Recognition Think of it as counting combinations of transitions between levels:

_nC_2

. For

n=4

_4C_2 = 6

lines. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

Q5 jee_main_2025_24_jan_morning Bohr Model of the Hydrogen Atom

During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is 2000 Å and it becomes 6000 Å when the electron jumps from state B to state C. Then the wavelength of the radiation emitted during the transition of electrons from state A to state B is :-

A. 3000 Å
B. 6000 Å
C. 4000 Å
D. 2000 Å

Solution

### Related Formula The energy of the emitted photon during an atomic transition between energy states is given by:

Delta E = frachclambda

where

h

is Planck's constant,

c

is speed of light, and

lambda

is the photon wavelength. ### Core Logic Write equations for the energy transitions from the layout of levels

Bohr Model of the Hydrogen Atom diagram for Q5 - JEE Main 2025 Morning

E_A - E_C = frachclambda_AC = frachc2000text AA quad dots (i)

E_B - E_C = frachclambda_BC = frachc6000text AA quad dots (ii)

### Step 1: Finding Transition A to B Subtracting equation

(ii)

from equation

(i)

gives the net transition energy from A to B :

E_A - E_B = (E_A - E_C) - (E_B - E_C)

frachclambda_AB = frachc2000 - frachc6000

frac1lambda_AB = frac3 - 16000 = frac26000 = frac13000

lambda_AB = 3000text AA

### Pattern Recognition Energy differences add linearly, which means their corresponding inverse wavelengths satisfy a parallel reciprocal subtraction rule:

frac1lambda_AB = frac1lambda_AC - frac1lambda_BC

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

Q7 jee_main_2025_28_jan_evening Bohr Model

The frequency of revolution of the electron in Bohr's orbit varies with

n

, the principal quantum number as

A. $frac1n$ $\frac{1}{n}$
B. $frac1mathrmn^3$ $\frac{1}{\mathrm{n}^3}$
C. $frac1mathrmn^4$ $\frac{1}{\mathrm{n}^4}$
D. $frac1n^2$ $\frac{1}{n^2}$

Solution

### Related Formula The orbital frequency of revolution

f

of an electron is inversely proportional to its time period

T

f = frac1T = fracv2pi r

In Bohr's Atomic Model: * Velocity

v propto fracZn

* Radius

r propto fracn^2Z

### Core Logic Substitute the proportional relationships of

v

and

r

into the frequency expression:

f propto fracleft(frac1nright)n^2 implies f propto frac1n^3

### Step 1: Verification Thus, the frequency varies inversely with the cube of the principal quantum number:

f propto frac1n^3

. ### Pattern Recognition Remember the sequence of powers of

n

in Bohr's model: radius expands as

n^2

, velocity drops as

n^-1

, angular momentum grows as

n^1

, and orbital time period or frequency changes as

n^3

n^-3

respectively. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

Q49 jee_main_2024_01_february_morning Hydrogen Spectrum

The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly:

A. 1.5 eV
B. 13.6 eV
C. 1.9 eV
D. 12.1 eV

Solution

### Related Formula Bohr state energy level values:

E_n = -frac13.6n^2mathrm~eV

Transition excitation requirement:

Delta E = E_textfinal - E_textinitial

### Core Logic To emit radiation in the Balmer series, the hydrogen electron must first be excited to at least the

n=3

shell. This allows it to jump down to

n=2

and produce the first spectral line of the Balmer series. Energy required to transition from ground state (

n=1

) to

n=3

E_1 = -13.6mathrm~eV

E_3 = -frac13.63^2 = -1.51mathrm~eV

### Step 1: Calculate Energy Gap

Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09mathrm~eV approx 12.1mathrm~eV

### Pattern Recognition Balmer emissions always return down to

n=2

. Thus, the initial excitation starting from ground state

n=1

must reach at least

n=3

to create a valid Balmer transition. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

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If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and $frac14pi in_0 = 9 times 10^9$ in SI units)

Solution & Explanation

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If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and frac14pi in_0 = 9 times 10^9$\frac{1}{4\pi \in_{0}} = 9 \times 10^{9}$ in SI units)

Solution & Explanation

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If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and $frac14pi in_0 = 9 times 10^9$ in SI units)