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 nth Bohr orbit is given by:
If you look closely at this expression, almost everything is a constant except for the principal quantum number n and the atomic number Z.
If we group all these constants together for the very first orbit of a hydrogen atom (where n=1 and Z=1), we get a magical number known as the Bohr radius, denoted by a0.
The value of a0 is approximately 0.529 A˚ (often rounded to 0.53 A˚ for quick calculations).
By substituting a0 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 Z=1.
We are asked to find the radius of the second Bohr orbit, so our principal quantum number n=2.
Let's substitute these simple values into our elegant formula:
Squaring the 2 gives us 4. Now, we just need to multiply:
And there we have it! The radius of the second Bohr orbit is exactly 2.12 A˚.
The Power of Proportionality
Before we move on, let's extract a deeper physical insight from this result. Notice the relationship rn∝n2.
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 9. If they had asked for a Helium ion (He+), you would divide by its atomic number Z=2.
Always keep these proportionalities at your fingertips. They are the ultimate cheat codes for solving complex atomic structure problems in seconds!