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

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

$KE = -E$ $\text{KE} = -E$ $PE = 2E$ $\text{PE} = 2E$

KE = (1)/(2) |PE|

Core Logic

In any allowed Bohr orbit (for any value of principal quantum number $n$ $n$), the relationship between kinetic energy (KE), potential energy (PE), and total energy (E) strictly obeys the virial theorem for a Coulombic force field.

|PE| = 2 \times KE

Step 1: Formulate the Ratio

The question asks for the ratio of the magnitude of KE to the magnitude of PE.

KE|PE| = (1)/(2)

This ratio is independent of the orbit state. Even though it is the $5th$ $5^{\text{th}}$ excited state, the ratio remains $(1)/(2)$ $\frac{1}{2}$.

Pattern Recognition

Energy relationships in Bohr orbits (and planetary motion): $K = -E = -U/2$ $K = -E = -U/2$. The state number ( $n$ $n$) is given solely as a distractor.

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 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 — 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 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 — 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 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 :