When a hydrogen atom going from n = 2 to n = 1 emits a photon, its recoil speed is (x/5) ~m / s. Where x = ______

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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

When a hydrogen atom going from n = 2$n = 2$ to n = 1$n = 1$ emits a photon, its recoil speed is (x)/(5) ~m / s$\frac{x}{5} \mathrm{~m / s}$. Where x =$x =$ _______. (Use: mass of hydrogen atom = 1.6 × 10⁻²⁷ ~kg$= 1.6 \times 10^{-27} \mathrm{~kg}$)

Solution & Explanation

Related Formula

Core Logic

Step 1: Compute Recoil Speed

Step 2: Simplify Arithmetic Fractions

Step 3: Solve for Variable x

Pattern Recognition

Chapter Mix

More Atoms Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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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When a hydrogen atom going from $n = 2$ $n = 2$ to $n = 1$ $n = 1$ emits a photon, its recoil speed is $(x)/(5) ~m / s$ $\frac{x}{5} \mathrm{~m / s}$. Where $x =$ $x =$ _______. (Use: mass of hydrogen atom $= 1.6 \times 10⁻²⁷ ~kg$ $= 1.6 \times 10^{-27} \mathrm{~kg}$)

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