Question 18

The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly :

Solution

Share

For Balmer series, the minimum energy transition is n=3 to n=2: E = 13.6(1/4-1/9) = 1.89 eV.

But to emit in Balmer series, atom must first reach n≥3 from ground state n=1: E = 13.6(1-1/9) = 12.09 ≈ 12.1 eV.

The correct answer is Option 4: 12.1 eV.

Create a FREE account and get:

  • Free JEE Mains Previous Papers PDF
  • Take JEE Mains paper tests