The Question is from neet-physics , and it was created by the author Mr. ANIL KUMAR.G

Question

Which of the following statements are true regarding Bohr’s model of hydrogen atom 
(I) Orbiting speed of electron decreases as it shifts to discrete orbits away from the nucleus 
(II) Radii of allowed orbits of electron are proportional to the principal quantum number 
(III) Frequency with which electrons orbit around the nucleus is discrete orbits is inversely proportional to the cube of principal quantum number 
(IV) Binding force with which the electron is bound to the nucleus increases as it shifts to outer orbits 
Select correct answer using the codes given below

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Bohr's model of the hydrogen atom
	**Related Topics**: Quantization of energy levels, orbital radius, electron velocity, orbital frequency, binding energy
	**Prerequisites**: Understanding of Bohr's postulates, relationship between energy levels and orbital parameters
## SOLUTION PATHWAY
**Given Information**:
	Statements regarding Bohr's model and hydrogen atom.
	Correct answer is I and III.
**Method & Steps**:
	1. Analyze statement (I):  The orbital velocity of an electron in a Bohr orbit is given by $v = \frac{e^2}{2\epsilon_0 nh}$, where n is the principal quantum number. Thus, as n increases (moving to outer orbits), the velocity decreases. Statement I is correct.
		Reasoning: The velocity of the electron is inversely proportional to the principal quantum number.
		Formula/Rule: $v = \frac{e^2}{2\epsilon_0 nh}$
	2. Analyze statement (II): The radius of the Bohr orbit is given by $r = \frac{\epsilon_0 n^2 h^2}{\pi m e^2}$. Therefore, the radius is proportional to the square of the principal quantum number (n²), not 'n' . Statement II is incorrect.
		Reasoning: Orbital radius is proportional to n²
		Formula/Rule: $r = \frac{\epsilon_0 n^2 h^2}{\pi m e^2}$
	3. Analyze statement (III): The frequency of revolution of the electron is $f = \frac{v}{2\pi r}$. Substituting the expressions for v and r, we find that frequency is proportional to $\frac{1}{n^3}$. Thus, the frequency is inversely proportional to the cube of the principal quantum number. Statement III is correct.
		Reasoning: Orbital frequency is inversely proportional to n³
		Formula/Rule: $f = \frac{v}{2\pi r}$
	4. Analyze statement (IV): The binding energy (or binding force) decreases as the electron goes to higher orbits because the energy levels become less negative as n increases. Statement IV is incorrect.
		Reasoning: Binding energy decreases in outer orbits.
		Key Point: Higher orbits have lower (less negative) binding energy.
	5. Conclusion: Statements I and III are correct, while II and IV are incorrect.
		Result: Only Statements I and III are correct.
		Verification:  The given answer (A) matches the analysis.
 ## OPTION ANALYSIS
[A] ✓: Statement I and III are correct.
[B] :x: Statement II and IV are incorrect.
[C] :x: Statement II is incorrect and IV is incorrect.
[D] :x: Statement II and IV are incorrect.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. Confusing proportionality with direct dependence: The radius is proportional to n², not n.
		2. Incorrectly assuming increasing binding force with increasing orbit radius: Binding energy decreases with increasing orbit radius.
	**Quick Tips**:
		1. Remember the key formulas for Bohr radius, velocity and frequency to verify proportionality statements.
		2. Understand the concept of binding energy and how it relates to the energy levels and orbit radius.

Mr. ANIL KUMAR.G · Answer

The correct Option is "I and III"

Mr. ANIL KUMAR.G · Answer

II and IV

Mr. ANIL KUMAR.G · Answer

I, II and III

Mr. ANIL KUMAR.G · Answer

II, III and IV

NEET Physics MCQs

Select Chapter

Question Tags

Question Tags

NEET Subjects

Our Bots

Company