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…

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More Atoms Previous-Year Questions

Q41 jee_main_2026_21_jan_evening Bohr Model

The energy of an electron in an orbit of the Bohr's atom is

-0.04E₀ eV

where

E₀

is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then

(2π L)/(h)

is :

A. $2$ $2$
B. $4$ $4$
C. $5$ $5$
D. $6$ $6$

Solution

Related Formula

Eₙ = (E₀)/(n²)

L = (nh)/(2π)

Core Logic

From Bohr's theory, the energy of an electron in the $n$ $n$-th orbit is inversely proportional to $n²$ $n^2$:

E = -(E₀)/(n²)

Equating this to the given energy:

-(E₀)/(n²) = -0.04 E₀

(1)/(n²) = 0.04 = (1)/(25)

n² = 25 n = 5

Step 1: Final Conclusion

Bohr's quantization condition for angular momentum:

L = (nh)/(2π)

Rearranging for the required term:

(2π L)/(h) = n

Since $n=5$ $n=5$, the value is $5$ $5$.

Pattern Recognition

The expression $2π L / h$ $2\pi L / h$ is a direct request for the principal quantum number $n$ $n$. $Eₙ = E₁ / n²$ $E_n = E_1 / n^2$ allows finding $n$ $n$ instantly.

Chapter Mix

Class 12 Physics: Atoms

Q32 jee_main_2026_22_january_evening Hydrogen Spectrum and Spectral Series

The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly ____ nm.

A. $1875$ $1875$
B. $1550$ $1550$
C. $1217$ $1217$
D. $1784$ $1784$

Solution

Related Formula

(1)/(λ) = R ( (1)/(n₁²) - (1)/(n₂²) )

Core Logic

For smallest wavelength of Lyman series ( $n₁ = 1, n₂ = \infty$ $n_1 = 1, n_2 = \infty$):

1λL, min = R ((1)/(1²) - 0) R = (1)/(91) ~nm⁻¹

For largest wavelength of Balmer series ( $n₁ = 2, n₂ = 3$ $n_1 = 2, n_2 = 3$):

1λB = R ((1)/(4) - (1)/(9)) = (1)/(91) \times (5)/(36) λB = (91 \times 36)/(5) = 655.2 ~nm

For largest wavelength of Paschen series ( $n₁ = 3, n₂ = 4$ $n_1 = 3, n_2 = 4$):

1λP = R ((1)/(9) - (1)/(16)) = (1)/(91) \times (7)/(144) λP = (91 \times 144)/(7) = 1872 ~nm

Difference in wavelengths:

Δ λ = λP - λB = 1872 - 655.2 = 1216.8 ~nm \approx 1217 ~nm

Step 1: Final Conclusion

The wavelength difference is approximately $1217 ~nm$ $1217 \mathrm{~nm}$.

Pattern Recognition

Spectral lines: Smallest wavelength $arrow n₂ = \infty$ $\rightarrow n_2 = \infty$. Largest wavelength $arrow n₂ = n₁ + 1$ $\rightarrow n_2 = n_1 + 1$. Rydberg constant $R = 1/λL,min$ $R = 1/\lambda_{L,\text{min}}$. Substitute values into Balmer ( $2arrow 3$ $2\rightarrow 3$) and Paschen ( $3arrow 4$ $3\rightarrow 4$) equations.

Chapter Mix

Class 12 Physics: Atoms

Q38 jee_main_2026_23_january_morning Hydrogen Spectrum

In hydrogen atom spectrum, (

R arrow

Rydberg's constant) A. the maximum wavelength of the radiation of Lyman series is

(4)/(3R)

B. the Balmer series lies in the visible region of the spectrum C. the minimum wavelength of the radiation of Paschen series is

(9)/(R)

D. the minimum wavelength of Lyman series is

(5)/(4R)

Choose the correct answer from the options given below:

A. B, D Only
B. A, B and C Only
C. A, B and D Only
D. A, B Only

Solution

Related Formula

(1)/(λ) = R( 1n₁² - 1n₂²)

Core Logic

Assess each statement using the Rydberg formula. Max wavelength corresponds to min energy (transition from $n₁+1 \to n₁$ $n_{1}+1 \to n_{1}$), and min wavelength corresponds to max energy ( $n₂ \to \infty$ $n_{2} \to \infty$).

Step 1: Evaluate A and D (Lyman Series)

For Lyman series, $n₁ = 1$ $n_{1} = 1$. Maximum wavelength ( $n₂ = 2$ $n_{2} = 2$):

1λ = R(1 - (1)/(4)) = (3R)/(4) λ = (4)/(3R)

Statement A is correct.

Minimum wavelength ( $n₂ = \infty$ $n_{2} = \infty$):

1λ = R(1 - 0) = R λ = (1)/(R)

Statement D is incorrect.

Step 2: Evaluate B (Balmer Series)

Balmer series corresponds to transitions to $n₁ = 2$ $n_{1} = 2$. These transitions primarily emit in the visible spectrum. Statement B is correct.

Step 3: Evaluate C (Paschen Series)

For Paschen series, $n₁ = 3$ $n_{1} = 3$. Minimum wavelength ( $n₂ = \infty$ $n_{2} = \infty$):

1λ = R((1)/(9) - 0) = (R)/(9) λ = (9)/(R)

Statement C is correct.

Step 4: Final Conclusion

Statements A, B, and C are correct.

Pattern Recognition

Sees: "minimum/maximum wavelength" → Max wavelength = adjacent orbital drop ( $n+1$ $n+1$ to $n$ $n$). Min wavelength = drop from infinity ( $\infty$ $\infty$ to $n$ $n$).

Chapter Mix

Class 12 Physics: Atoms

Q32 jee_main_2026_24_january_morning Bohr Model

Two electrons are moving in orbits of two hydrogen like atoms with speeds

3 \times 10⁵ m/s

and

2.5 \times 10⁵ m/s

respectively. If the radii of these orbits are nearly same then the possible order of energy states are ____ respectively.

Bohr orbits speed and radius relations diagram — Mathematical proportionality of velocity and radius in Bohr's model.

A. $6 and 5$ $\text{6 and 5}$
B. $9 and 8$ $\text{9 and 8}$
C. $8 and 10$ $\text{8 and 10}$
D. $10 and 12$ $\text{10 and 12}$

Solution

Related Formula

v \propto (Z)/(n)

r \propto (n²)/(Z)

Core Logic

From the proportionalities of speed and radius in the Bohr model:

r \propto (n²)/(Z)

Since $v \propto (Z)/(n)$ $v \propto \frac{Z}{n}$, we can substitute $Z \propto nv$ $Z \propto nv$. This gives $r \propto (n²)/(nv) = (n)/(v)$ .

Step 1: Ratio Analysis

If the radii are the same, then:

(n₁)/(v₁) = (n₂)/(v₂)

(n₁)/(n₂) = (v₁)/(v₂) = (3 \times 10⁵)/(2.5 \times 10⁵) = (3)/(2.5) = (6)/(5)

Thus, the possible order of energy states ( $n$ $n$) is 6 and 5.

Pattern Recognition

By coupling $r$ $r$ and $v$ $v$ dependencies on $Z$ $Z$ and $n$ $n$, the $Z$ $Z$ cancels out allowing a direct ratio linking radius, orbit number, and velocity.

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 × 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

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate A and D (Lyman Series)

Step 2: Evaluate B (Balmer Series)

Step 3: Evaluate C (Paschen Series)

Step 4: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Ratio Analysis

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 $14π ε₀ = 9 \times 10⁹$ in SI units)