Single Choice

The wavelength of third line of the Balmer series for a H atom is:

A21/100R
B100/21R
Correct Answer
C21R/100
D100R/21

Solution

Balmer series $$\Rightarrow {n}_{1}=2$$

The third line of Balmer series $$\Rightarrow {n}_{2}=5$$

Rydberg equation $$\Rightarrow \cfrac {1} {\lambda}= {Rz}^{2} \left[ \cfrac {1} {{n}_{1}^{2}} - \cfrac {1} {{n}_{2}^{2}} \right]$$

$$\Rightarrow \cfrac {1}{\lambda } = \left( 10967800 \right) \left(1 \right)^{2} $$

$$\left[ \cfrac { 1 }{ { { 2 }^{ 2 } }} -\cfrac { 1 }{ { 5 }^{ 2 } } \right] $$

$$ \cfrac {1}{\lambda }= \cfrac {21} {100}R$$
Wavelength of $${3}^{rd}$$ line of Balmer, $$ \lambda = \cfrac {100} {21R}$$

So, the correct option is $$B$$


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