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 (2pi L/h) is :

From Bohr's theory, the energy of an electron in the n-th orbit is inversely proportional to n2: Equating this to the given energy: Bohr's quantization condition for angular momentum: Rearranging for the required term: Since n=5, the value is 5. The expression 2pi L / h is a direct request for the principal quantum number n. En = E1 / n2 allows finding n instantly. Class 12 Physics: Atoms

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, t

From Bohr's theory, the energy of an electron in the n-th orbit is inversely proportional to n2: Equating this to the given energy: Bohr's quantization condition for angular momentum: Rearranging for the required term: Since n=5, the value is 5. The expression 2pi L / h is a direct request for the principal quantum number n. En = E1 / n2 allows finding n instantly. Class 12 Physics: Atoms

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 (2pi L/h) is :

From Bohr's theory, the energy of an electron in the n-th orbit is inversely proportional to n2: Equating this to the given energy: Bohr's quantization condition for angular momentum: Rearranging for the required term: Since n=5, the value is 5. The expression 2pi L / h is a direct request for the principal quantum number n. En = E1 / n2 allows finding n instantly. Class 12 Physics: Atoms

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, t

From Bohr's theory, the energy of an electron in the n-th orbit is inversely proportional to n2: Equating this to the given energy: Bohr's quantization condition for angular momentum: Rearranging for the required term: Since n=5, the value is 5. The expression 2pi L / h is a direct request for the principal quantum number n. En = E1 / n2 allows finding n instantly. Class 12 Physics: Atoms

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…

Year	2026	2025	2024	Total
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Year

2026

2025

2024

Total

Questions

Solution & Explanation

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

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

The energy of an electron in an orbit of the Bohr's atom is $-0.04E₀ eV$ $-0.04E_{0} \text{ eV}$ where $E₀$ $E_{0}$ 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 $(2π L)/(h)$ $\frac{2\pi L}{h}$ is :

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

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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The energy of an electron in an orbit of the Bohr's atom is -0.04E₀ eV$-0.04E_{0} \text{ eV}$ where E₀$E_{0}$ 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)$\frac{2\pi L}{h}$ is :

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

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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The energy of an electron in an orbit of the Bohr's atom is $-0.04E₀ eV$ $-0.04E_{0} \text{ eV}$ where $E₀$ $E_{0}$ 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 $(2π L)/(h)$ $\frac{2\pi L}{h}$ is :