Question

The value of wavelength of radiation emitted due to transition of electrons from $n$=4 to $n$=2 state in hydrogen atom will be

• A. $\frac{5R}{36}$
• B. $\frac{16}{3R}$
• C. $\frac{36}{5R}$
• D. $\frac{3R}{16}$

Answer 1

Answer: (B)

$\frac{16}{3R}$

We know that, $\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

For hydrogen atom, atomic no. $Z=1$

So, $\frac{1}{\lambda }=R({\frac{1}{2}}^2-{\frac{1}{4}}^2)=\frac{3R}{16}$

$\Rightarrow \lambda =\frac{16}{3R}$