Considering the Bohr model of hydrogen like atoms, the ratio of the radius 5th orbit of the electron in Li²⁺ and He⁺ is

Solution & Explanation

Related Formula

Bohr orbit radius expression:

rₙ ∝ (n²)/(Z)

For a constant orbit index (n = 5):

r₅ ∝ (1)/(Z)

where Z is the atomic atomic number identifier.

Core Logic

Identify the atomic values:

  • For Li²⁺ implies ZLi = 3
  • For He⁺ implies ZHe = 2
Step 1: Compute Ratio

Set up the inverse scaling ratio format:

rLi²⁺rHe⁺ = ZHeZLi = (2)/(3)
Pattern Recognition

Since orbit numbers are identical, the radius is purely inversely proportional to the atomic nuclear charge Z.

Chapter Mix

Class 12 Physics: Atoms

Reference Study Guides

More Atomic Structure Previous-Year Questions — Page 5

Q45 jee_main_2024_31_jan_evening Nuclear Radius
The mass number of nucleus having radius equal to half of the radius of nucleus with mass number 192 is:
  • A. 24
  • B. 32
  • C. 40
  • D. 20

Solution

Related Formula

Nuclear radius is empirically related to mass number by: R = R₀ A1/3

Core Logic

Given R₁ = (R₂)/(2) where A₂ = 192. We need to find A₁.

Step 1: Forming the Ratio
(R₁)/(R₂) = ((A₁)/(A₂))1/3 (1)/(2) = ((A₁)/(192))1/3
Step 2: Cubing Both Sides
((1)/(2))³ = (A₁)/(192) (1)/(8) = (A₁)/(192) A₁ = (192)/(8) = 24
Pattern Recognition

Since R ∝ A1/3, scaling R by k means scaling A by k³. Half the radius (k = 1/2) means 1/8th the mass number.

Chapter Mix

Class 12 Physics: Nuclei

Q60 jee_main_2024_31_jan_evening Nuclear Size and Density
A nucleus has mass number A₁ and volume V₁. Another nucleus has mass number A₂ and volume V₂. If relation between mass number is A₂ = 4A₁, then (V₂)/(V₁) = ________.
Numerical Answer. Answer: 4 to 4

Solution

Related Formula

R = R₀ A1/3

V = (4)/(3)π R³
Core Logic

Since radius R is proportional to A1/3, the volume V (which depends on R³) will be directly proportional to the mass number A.

Step 1: Show Proportionality
V = (4)/(3)π (R₀ A1/3)³ = (4)/(3)π R₀³ A

This proves that V ∝ A.

Step 2: Calculate Ratio
(V₂)/(V₁) = (A₂)/(A₁)

Given that A₂ = 4A₁:

(V₂)/(V₁) = (4A₁)/(A₁) = 4
Pattern Recognition

Nuclear density is constant for all nuclei. Therefore, Mass ∝ Volume. Since Mass number (A) represents mass, Volume is strictly linearly proportional to Mass number.

Chapter Mix

Class 12 Physics: Nuclei

Q60 jee_main_2024_31_jan_morning Mass Defect And Energy
The mass defect in a particular reaction is 0.4 g. The amount of energy liberated is n × 10⁷ kWh where n = _______. (speed of light = 3 × 10⁸ m/s)
Numerical Answer. Answer: 1 to 1

Solution

Related Formula
E = Δ m c² 1 kWh = 3.6 × 10⁶ J
Core Logic

Given the mass defect:

Δ m = 0.4 g = 0.4 × 10⁻³ kg

The total energy liberated in Joules is:

E = (0.4 × 10⁻³) × (3 × 10⁸)² E = 0.4 × 10⁻³ × 9 × 10¹⁶ E = 3.6 × 10¹³ J
Step 2: Conversion to kWh

We need the answer in kWh. Since 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J:

E = 3.6 × 10¹³3.6 × 10⁶ kWh E = 10⁷ kWh

Comparing this to n × 10⁷ kWh, we get: n = 1

Chapter Mix

Class 12 Physics: Nuclei

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