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 epsilon0) = 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 epsilon0) = 9 x 109 in SI un

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 epsilon0) = 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 epsilon0) = 9 x 109 in SI un

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…

Year	2026	2025	2024	Total
Questions	5	8	9	22

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Kinitial = Uclosest approach

K = 14π ε₀ ((2e)(Ze))/(r₀)

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 ( $r₀$ $r_0$). $Kᵢ + Uᵢ = Kf + Uf$ $K_i + U_i = K_f + U_f$

Kᵢ + 0 = 0 + (1)/(4πε₀) ((2e)(79e))/(r₀)

Step 1: Convert Energy and Solve

Initial kinetic energy $Kᵢ = 7.7 MeV = 7.7 \times 10⁶ \times 1.6 \times 10⁻¹⁹ J$ .

7.7 \times 10⁶ \times 1.6 \times 10⁻¹⁹ = 9 \times 10⁹ \times (2 \times 1.6 \times 10⁻¹⁹) \times (79 \times 1.6 \times 10⁻¹⁹)r₀

r₀ = 9 \times 10⁹ \times 2 \times 79 \times (1.6 \times 10⁻¹⁹)²7.7 \times 10⁶ \times 1.6 \times 10⁻¹⁹

r₀ = 9 \times 10⁹ \times 158 \times 1.6 \times 10⁻¹⁹7.7 \times 10⁶

r₀ = 2275.2 \times 10⁻¹⁰7.7 \times 10⁶ = 295.48 \times 10⁻¹⁶ m \approx 2.95 \times 10⁻¹⁴ m

Pattern Recognition

Distance of closest approach problem: Simply equate initial Kinetic Energy (in Joules) to Potential Energy $k(Z₁e)(Z₂e)/r₀$ $k(Z_1e)(Z_2e)/r_0$. Alpha particle has charge $2e$ $2e$.

Chapter Mix

Class 12 Physics: Atoms

More Atoms Previous-Year Questions — Page 5

Q53 jee_main_2024_30_jan_morning Hydrogen Energy Levels and Transitions

A electron of hydrogen atom on an excited state is having energy

Eₙ = -0.85 ~eV

. The maximum number of allowed transitions to lower energy level is ....

Numerical Answer. Answer: 6 to 6

Solution

Related Formula

Eₙ = -(13.6)/(n²) ~eV

Number of transitions = (n(n - 1))/(2)

Core Logic

First, identify the principal quantum number $n$ $n$ corresponding to the energy $-0.85 ~eV$ $-0.85 \mathrm{~eV}$. Then, use the combinatorics formula to find the total possible downward emission transitions.

Step 1: Find Quantum State

Eₙ = -(13.6)/(n²)

-0.85 = -(13.6)/(n²)

n² = (13.6)/(0.85) = 16

$n = 4$ $n = 4$

Step 2: Calculate Transitions

Maximum number of transitions from $n=4$ $n=4$ to lower levels ( $n=3, 2, 1$ $n=3, 2, 1$):

= (n(n - 1))/(2)

= (4(4 - 1))/(2) = (12)/(2) = 6

Pattern Recognition

Energy states in Hydrogen are heavily standardized: $n=1 arrow -13.6$ $n=1 \rightarrow -13.6$, $n=2 arrow -3.4$ $n=2 \rightarrow -3.4$, $n=3 arrow -1.51$ $n=3 \rightarrow -1.51$, $n=4 arrow -0.85$ $n=4 \rightarrow -0.85$. Recognize $-0.85 ~eV$ $-0.85 \mathrm{~eV}$ as state 4 immediately.

Chapter Mix

Class 12 Physics: Atoms

Q36 jee_main_2024_31_jan_morning Hydrogen Spectrum

If the wavelength of the first member of Lyman series of hydrogen is

λ

. The wavelength of the second member will be

A. $(27)/(32) λ$ $\frac{27}{32} \lambda$
B. $(32)/(27)λ$ $\frac{32}{27}\lambda$
C. $(27)/(5) λ$ $\frac{27}{5} \lambda$
D. $(5)/(27)λ$ $\frac{5}{27}\lambda$

Solution

Related Formula

(1)/(λ) = R Z² [ (1)/(n₁²) - (1)/(n₂²) ]

Core Logic

For the first member of the Lyman series of hydrogen ( $n₁ = 1, n₂ = 2$ $n_1 = 1, n_2 = 2$):

(1)/(λ) = (13.6 Z²)/(hc) [ (1)/(1²) - (1)/(2²) ]

(1)/(λ) = (13.6 Z²)/(hc) [ (3)/(4) ] (i)

Step 2: Second Member Calculation

For the second member of the Lyman series ( $n₁ = 1, n₂ = 3$ $n_1 = 1, n_2 = 3$):

(1)/(λ') = (13.6 Z²)/(hc) [ (1)/(1²) - (1)/(3²) ]

(1)/(λ') = (13.6 Z²)/(hc) [ (8)/(9) ] (ii)

Step 3: Ratio

On dividing equation (i) by (ii):

(λ')/(λ) = (3/4)/(8/9) = (3)/(4) \times (9)/(8)

(λ')/(λ) = (27)/(32)

λ' = (27)/(32) λ

Pattern Recognition

Rydberg ratios between members of the same series are purely derived from the bracket terms $[1/n₁² - 1/n₂²]$ $[1/n_1^2 - 1/n_2^2]$. For Lyman 1st and 2nd, the ratio is $(3/4) / (8/9) = 27/32$ $(3/4) / (8/9) = 27/32$.

Chapter Mix

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 $14π ε₀ = 9 \times 10⁹$ in SI units)

Solution & Explanation

Related Formula

Core Logic

Step 1: Convert Energy and Solve

Pattern Recognition

Chapter Mix

Reference Study Guides

More Atoms Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Find Quantum State

Step 2: Calculate Transitions

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Second Member Calculation

Step 3: Ratio

Pattern Recognition

Chapter Mix

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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 14π ε₀ = 9 × 10⁹$\frac{1}{4\pi \epsilon_{0}} = 9 \times 10^{9}$ in SI units)

Solution & Explanation

Related Formula

Core Logic

Step 1: Convert Energy and Solve

Pattern Recognition

Chapter Mix

Reference Study Guides

More Atoms Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Find Quantum State

Step 2: Calculate Transitions

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Second Member Calculation

Step 3: Ratio

Pattern Recognition

Chapter Mix

More Atoms Questions — jee_main_2026_21_jan_morning

Select Your Target Exam

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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 $14π ε₀ = 9 \times 10⁹$ in SI units)