Question

The Third Line in Balmer series correspondence to an electronic transition between which Bohr's orbits in hydrogen:

• A. $5\to 3$
• B. $5\to 2$
• C. $4\to 3$
• D. $4\to 2$

Answer 1

Answer: (B)

$5\to 2$

When an electron comes down from higher energy level to second energy level, then Balmer series of the spectrum is obtained.

Wavelength of $H$-atom is given by,

$\frac{1}{\lambda }=R\times [\frac{1}{2^2}-\frac{1}{n_2^2}]$

The first line is $3\to 2$, second line is $4\to 2$ and third line is $5\to 2$.

Hence, the correct option is $B$