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

Year	2026	2025	2024	Total
Questions	5	8	9	22

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

By conservation of linear momentum, the momentum of the recoiling hydrogen atom matches the momentum of the emitted photon:

patom = pphoton m v = (Δ E)/(c)

Hence, the recoil speed $v$ $v$ is:

v = (Δ E)/(m c)

where $Δ E$ $\Delta E$ is the energy difference between the electronic energy levels.

Core Logic

For a hydrogen transition from $n = 2$ $n = 2$ to $n = 1$ $n = 1$:

Δ E = E₂ - E₁ = -3.4 ~eV - (-13.6 ~eV) = 10.2 ~eV

Converting energy to Joules:

Δ E = 10.2 \times 1.6 \times 10⁻¹⁹ ~J

Electronic transition energy level diagram from n=2 to n=1 for Q51

Step 1: Compute Recoil Speed

Substituting given constant values:

v = 10.2 \times 1.6 \times 10⁻¹⁹ ~J(1.6 \times 10⁻²⁷ ~kg) \times (3 \times 10⁸ ~m/s)

Step 2: Simplify Arithmetic Fractions

Cancelling out the factor $1.6$ $1.6$ from numerator and denominator:

v = 10.2 \times 10⁻¹⁹3 \times 10⁻¹⁹ = (10.2)/(3) = 3.4 ~m/s

Step 3: Solve for Variable x

Equating the computed speed to the given parameter form:

3.4 = (x)/(5) x = 3.4 \times 5 = 17

Therefore, the value of $x$ $x$ is $17$ $17$.

Pattern Recognition

Always look for clean component cancellation matches prior to raw exponential expansion. In this problem, the value $1.6$ $1.6$ in the electron charge value cancels with the structural atomic mass constant value perfectly.

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

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 $= 1.6 \times 10⁻²⁷ ~kg$ $= 1.6 \times 10^{-27} \mathrm{~kg}$)

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 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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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 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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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 $= 1.6 \times 10⁻²⁷ ~kg$ $= 1.6 \times 10^{-27} \mathrm{~kg}$)