QJEEM17Che378: JEE Main 2017 Chemistry - Class 11 JEEM17Che378 Solution: The electron in the hydrogen atom undergoes transition from…

Here electron is coming to the orbital of radius 211.6 pm. Now we have to find which series have radius of orbital 211.6 pm.

We know,

Radius, r = $$0.529\times \frac{n^2}{Z}\stackrel{\circ }{A}$$

Given, r = 211.6 pm = 211.6 $$\times$$ 10^-12 m

and Z = 1 for hydrogen atom

$$\therefore$$ 211.6 $$\times$$ 10^-12 = $$0.529\times \frac{n^2}{1}$$ $$\times$$ 10^-10

$$\Rightarrow$$ n = 2

As n = 2 so the series is Balmer Series.