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

Q11 jee_main_2025_02_april_evening Bohr's Atomic Model

Assuming the validity of Bohr's atomic model for hydrogen like ions the radius of

Li⁺⁺

ion in its ground state is given by

1Xa₀

, where

X = _ _ _ .

(Where

a₀

is the first Bohr's radius.)

A. $2$ $2$
B. $1$ $1$
C. $3$ $3$
D. $9$ $9$

Solution

Related Formula

Bohr radius formula for hydrogen-like species:

rₙ = a₀ (n²)/(Z)

Core Logic

For Lithium ion $Li⁺⁺$ $\mathrm{Li}^{++}$:

Atomic number $Z = 3$ $Z = 3$
For ground state, the principal quantum number $n = 1$ $n = 1$

Substitute $Z = 3$ $Z = 3$ and $n = 1$ $n = 1$ into Bohr's radius formula:

r₁ = a₀ (1²)/(3) = (a₀)/(3)

Comparing with the given expression $(1)/(X) a₀$ $\frac{1}{X} a_0$:

(1)/(X) a₀ = (a₀)/(3) X = 3

Pattern Recognition

Sees: Ground state radius of hydrogen-like species. Trap: Confusing the atomic number $Z$ $Z$ of Lithium with Helium ( $Z=2$ $Z=2$) or Beryllium ( $Z=4$ $Z=4$). Shortcut: Ground state radius of hydrogenic species is simply $a₀ / Z$ $a_0 / Z$. Since Lithium has $Z=3$ $Z=3$, the ground state radius must be $a₀/3$ $a_0/3$ directly.

Chapter Mix

Class 12 Physics: Atoms

Q7 jee_main_2025_03_april_evening Bohr's Model of Hydrogen Atom

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The Bohr model is applicable to hydrogen and hydrogen-like atoms only. Reason R: The formulation of Bohr model does not include repulsive force between electrons. In the light of the above statements, choose the correct answer from the options given below :

A. Both A and R are true but R is NOT the correct explanation of A.
B. A is false but R is true.
C. Both A and R are true and R is the correct explanation of A.
D. A is true but R is false.

Solution

Related Formula

Bohr's electrostatic centripetal balance equation:

(m v²)/(r) = (1)/(4πε₀) (Z e²)/(r²)

This basic equation matches ONLY a single electron orbiting a nucleus of charge $+Ze$ $+Ze$.

Core Logic

Assertion Analysis:

Bohr's model matches single-electron species (such as $H$ $\mathrm{H}$, $He^+$ $\mathrm{He}^+$, $Li²⁺$ $\mathrm{Li}^{2+}$, etc.). Hence, Assertion A is true.

Reason Analysis:

The model is restricted because it models ONLY the attractive force between the positive nucleus and one orbiting electron. It cannot handle multi-electron systems due to the presence of inter-electron repulsive forces, which are not integrated into Bohr's simple formulation. Hence, Reason R is true.

Connection Check:

The lack of repulsive forces is exactly why the model fails for multi-electron species and remains applicable only to single-electron (hydrogen-like) species. Thus, R is the correct explanation of A.

Pattern Recognition

For single vs. multi-electron systems in atomic physics: Bohr model = single-electron ONLY. Quantum mechanics (Schrodinger) is required for multi-electron systems due to electron-electron interactions.

Chapter Mix

Class 12 Physics: Atoms

Q24 jee_main_2025_03_april_evening Hydrogen Spectrum and Energy Level Transitions

An electron in the hydrogen atom initially in the fourth excited state makes a transition to

nth

energy state by emitting a photon of energy 2.86 eV. The integer value of n will be ________.

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

The energy of an electron in the $n$ $n$-th shell of a hydrogen atom is:

Eₙ = -(13.6)/(n²)~eV

The emitted photon energy during transition $nᵢ arrow nf$ $n_i \rightarrow n_f$ is:

Δ E = Enᵢ - Enf

Core Logic Given state: Initial state: "fourth excited state"$ Core Logic Given state: Initial state: "fourth excited state" $\Rightarrow n_i = 5 Emitted photon energy:$ Emitted photon energy: $\Delta E = 2.86\mathrm{~eV} Step 1: Calculate energy of initial state ($ Step 1: Calculate energy of initial state ($E_5)

E₅ = -(13.6)/(5²) = -(13.6)/(25) = -0.544~eV

Step 2: Calculate final state energy ($Eₙ$)

2.86 = E_5 - E_n = -0.544 - E_n

E_n = -0.544 - 2.86 = -3.404\mathrm{~eV}

$ Step 3: Determine the integer shell index ($n$)

-3.4 = -\frac{13.6}{n^2}

n^2 = \frac{13.6}{3.4} = 4 \Rightarrow n = 2$

Pattern Recognition Memorizing the first few energy levels of the hydrogen atom saves calculation time: $ Pattern Recognition Memorizing the first few energy levels of the hydrogen atom saves calculation time: $E_1 = -13.6\mathrm{~eV} $ $E_2 = -3.4\mathrm{~eV} $ $E_3 = -1.51\mathrm{~eV} $ $E_4 = -0.85\mathrm{~eV} $ $E_5 = -0.54\mathrm{~eV} Recognizing that a transition from$ Recognizing that a transition from $-0.54\mathrm{~eV}by emitting$ by emitting $2.86\mathrm{~eV}lands exactly at$ lands exactly at $-3.4\mathrm{~eV}maps directly to$ maps directly to $n=2$ instantly. Chapter Mix Class 12 Physics: Atoms

Q15 jee_main_2025_07_april_morning Bohr Model of Hydrogen Atom

In a hydrogen like ion, the energy difference between the

2nd

excitation energy state and ground is

108.8eV

. The atomic number of the ion is

A. 4
B. 2
C. 1
D. 3

Solution

Related Formula

For a hydrogen-like ion of atomic number $Z$ $Z$, the transition energy between states $n₂$ $n_2$ and $n₁$ $n_1$ is:

Δ E = 13.6 Z² ( (1)/(n₁²) - (1)/(n₂²) ) ~eV

Core Logic Identify the states: Ground state:$ Core Logic Identify the states: Ground state: $n_1 = 1 $ $2^{\mathrm{nd}}excitation energy state:$ excitation energy state: $n_2 = 3 Substitute values into the energy equation: $ Substitute values into the energy equation: $

108.8 = 13.6 Z² ( (1)/(1²) - (1)/(3²) )

108.8 = 13.6 Z² ( 1 - (1)/(9) ) = 13.6 Z² \times (8)/(9)

Step 1: Solve for Z Rearrange the equation to isolate $ Step 1: Solve for Z Rearrange the equation to isolate $Z^2 :

Z² = (108.8 \times 9)/(13.6 \times 8)

Calculate the numerical value:

(108.8)/(13.6) = 8

Z² = (8 \times 9)/(8) = 9 Z = 3

Pattern Recognition Sees: "$ Pattern Recognition Sees: "$2^{\mathrm{nd}}excitation"$ excitation" $\implies n = 3. Ground state$. Ground state $\implies n = 1. Shortcut: Energy transition ratio for$. Shortcut: Energy transition ratio for $1 \to 3is always$ is always $\frac{8}{9}of$ of $13.6 Z^2 \approx 12.1 Z^2. Since$. Since $108.8 / 12.1 \approx 9,$, $Z^2 = 9 \implies Z = 3$. Chapter Mix Class 12 Physics: Atoms

Q16 jee_main_2025_07_april_morning Hydrogen Spectrum

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is.

A. $5:36$ $5:36$
B. $5:27$ $5:27$
C. $3 : 4$ $3 : 4$
D. $27:5$ $27:5$

Solution

Related Formula

The wavelength $λ$ $\lambda$ for a transition in a hydrogen atom is given by the Rydberg formula:

(1)/(λ) = R ( (1)/(n₁²) - (1)/(n₂²) )

To find the largest wavelength (minimum energy transition), select the adjacent higher shell $n₂ = n₁ + 1$ $n_2 = n_1 + 1$.

Core Logic

Lyman Series largest wavelength ( $λL$ $\lambda_L$): Transition from $n = 2 \to 1$ $n = 2 \to 1$

(1)/(λL) = R ( (1)/(1²) - (1)/(2²) ) = (3R)/(4) λL = (4)/(3R)

Balmer Series largest wavelength ( $λB$ $\lambda_B$): Transition from $n = 3 \to 2$ $n = 3 \to 2$

(1)/(λB) = R ( (1)/(2²) - (1)/(3²) ) = R ( (1)/(4) - (1)/(9) ) = (5R)/(36) λB = (36)/(5R)

Step 1: Ratio Calculation

Now, compute the ratio of the wavelengths:

(λL)/(λB) = ((4)/(3R))/((36)/(5R)) = (4)/(3) \times (5)/(36) = (5)/(27)

Pattern Recognition Sees: Ratio of largest wavelengths of series. Shortcut: The largest wavelength in a series starting at ground level$ Pattern Recognition Sees: Ratio of largest wavelengths of series. Shortcut: The largest wavelength in a series starting at ground level $n_1is$ is $\lambda \propto \frac{n_1^2 (n_1+1)^2}{2n_1 + 1}. For Lyman ($. For Lyman ($n_1=1):$): $\lambda_L \propto \frac{4}{3}. For Balmer ($. For Balmer ($n_1=2):$): $\lambda_B \propto \frac{36}{5}. Ratio:$. Ratio: $\frac{4/3}{36/5} = \frac{5}{27}$. 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 2

Solution

Related Formula

Core Logic

Pattern Recognition

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Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Core Logic

Step 1: Calculate energy of initial state ($

Step 1: Calculate energy of initial state ($E_5)

Step 2: Calculate final state energy ($Eₙ$)

Step 2: Calculate final state energy ($E_n$)

Step 3: Determine the integer shell index ($n$)

Step 3: Determine the integer shell index ($n$)

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Core Logic

Step 1: Solve for Z

Step 1: Solve for Z

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Ratio Calculation

Pattern Recognition

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 :

Step 1: Calculate energy of initial state ($E_5 $)$