Sigma Percentile
JEE Main 2017
LEVELJEE Main

Animated Solution for Chemistry - Atomic Structure: The radius of the second Bohr orbit for hydrogen atom is (Planck's constant ; mass of electron ; charge of electron ; permitivity of vacuum )

Select Answer:

Visualized Solution

  • The hydrogen atom consists of a central nucleus and an electron revolving in circular orbits.
  • We need to find the radius of the second orbit ().

  • The general formula for the radius of the Bohr orbit is:
  • This complex expression evaluates to a simple constant for the first orbit of hydrogen, known as the Bohr radius, .

  • The Bohr radius is approximately (often taken as ).
  • The simplified formula becomes:

  • For a hydrogen atom, the atomic number .
  • We need the radius of the second orbit, so .
  • Substituting these values:

  • Thus, the radius of the second Bohr orbit is .

  • Notice that .
  • If we were asked for the 3rd orbit, .
  • For hydrogen-like ions (e.g., ), .

The Sigma Insight: Bohr's Model

Solution Diagram

The Intimidating Wall of Constants

When you first read this question, your heart might skip a beat. The problem throws a massive wall of fundamental constants at you: Planck's constant, the mass of an electron, the charge of an electron, and the permittivity of free space.
It almost feels like a trap, daring you to plug all these microscopic numbers into your calculator and waste precious minutes during the exam.
But as a smart JEE aspirant, you must recognize this classic misdirection. You don't need to calculate everything from scratch.

The Elegance of the Bohr Radius

The general formula for the radius of the Bohr orbit is given by:
If you look closely at this expression, almost everything is a constant except for the principal quantum number and the atomic number .
If we group all these constants together for the very first orbit of a hydrogen atom (where and ), we get a magical number known as the Bohr radius, denoted by .
The value of is approximately (often rounded to for quick calculations).
By substituting back into our original equation, the terrifying formula simplifies beautifully to:

Executing the Final Strike

Now, the problem becomes incredibly straightforward. We are dealing with a hydrogen atom, which means the atomic number .
We are asked to find the radius of the second Bohr orbit, so our principal quantum number .
Let's substitute these simple values into our elegant formula:
Squaring the gives us . Now, we just need to multiply:
And there we have it! The radius of the second Bohr orbit is exactly .

The Power of Proportionality

Before we move on, let's extract a deeper physical insight from this result. Notice the relationship .
This tells us that the orbits are not equally spaced. As you move further away from the nucleus, the gaps between successive orbits grow exponentially larger.
If the examiner had asked for the third orbit, you would simply multiply by . If they had asked for a Helium ion (), you would divide by its atomic number .
Always keep these proportionalities at your fingertips. They are the ultimate cheat codes for solving complex atomic structure problems in seconds!

Similar Questions

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The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to . The value of is ……… . ( is radius of Bohr's orbit) (Nearest integer) [Given, ]

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Comprehension Passage

Consider the Bohr's model of a one-electron atom where the electron moves around the nucleus. In the following List-I contains some quantities for the orbit of the atom and List-II contains options showing how they depend on . \begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\ \text{(I) Radius of the } n^{\text{th}} \text{ orbit} & \text{(P) } \propto n^{-2} \\ \text{(II) Angular momentum of the electron in the } n^{\text{th}} \text{ orbit} & \text{(Q) } \propto n^{-1} \\ \text{(III) Kinetic energy of the electron in the } n^{\text{th}} \text{ orbit} & \text{(R) } \propto n^0 \\ \text{(IV) Potential energy of the electron in the } n^{\text{th}} \text{ orbit} & \text{(S) } \propto n^1 \\ & \text{(T) } \propto n^2 \\ & \text{(U) } \propto n^{1/2} \end{array}
Question 1:

Which of the following options has the correct combination considering List-I and List-II ?

(A)
(II), (R)
(B)
(I), (P)
(C)
(I), (T)
(D)
(II), (Q)
Question 2:

Which of the following options has the correct combination considering List-I and List-II ?

(A)
(III), (S)
(B)
(IV), (Q)
(C)
(IV), (U)
(D)
(III), (P)
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According to Bohr's atomic theory, I. kinetic energy of electron is II. the product of velocity () of electron and principal quantum number (), '' . III. frequency of revolution of electron in an orbit is IV. coulombic force of attraction on the electron is Choose the most appropriate answer from the options given below.

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Only III
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I, III and IV
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Which of the following is the energy of a possible excited state of hydrogen?

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Given below are two statements. Statement I: According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in positive charges on the nucleus as there is no strong hold on the electron by the nucleus. Statement II: According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in principal quantum number. In the light of the above statements, choose the most appropriate answer from the options given below.

(A)
Both statement I and statement II are false.
(B)
Both statement I and statement II are true.
(C)
Statement I is false but statement II is true.
(D)
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