The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the 5th excited state of a…

Q: 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 :

KE = -E PE = 2E In any allowed Bohr orbit (for any value of principal quantum number n), the relationship between kinetic energy (KE), potential energy (PE), and total energy (E) strictly obeys the virial theorem for a Coulombic force field. The question asks for the ratio of the magnitude of KE to the magnitude of PE. This ratio is independent of the orbit state. Even though it is the 5th excited state, the ratio remains (1/2). Energy relationships in Bohr orbits (and planetary motion): K = -E

Year	2026	2025	2024	Total
Questions	5	8	9	22

Q38 jee_main_2026_21_jan_morning Alpha Particle Scattering

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 in₀ = 9 \times 10⁹

in SI units)

A. $2.95 \times 10⁻¹⁴$ $2.95 \times 10^{-14}$
B. $2.95 \times 10⁻¹⁶$ $2.95 \times 10^{-16}$
C. $3.85 \times 10⁻¹⁶$ $3.85 \times 10^{-16}$
D. $3.85 \times 10⁻¹⁴$ $3.85 \times 10^{-14}$

Solution

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

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

The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the 5th$5^{\text{th}}$ excited state of a hydrogen atom is :

Solution & Explanation

Related Formula

Core Logic

Step 1: Formulate the Ratio

Pattern Recognition

Chapter Mix

Reference Study Guides

More Atoms Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Convert Energy and Solve

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

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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

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The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the $5th$ $5^{\text{th}}$ excited state of a hydrogen atom is :