The ratio of minimum to maximum wavelength of radiation emitted by transition of an electron, to ground state of Bohr's hydrogen atom is,

The correct option is A $\frac{3}{4}$
For minimum wavelength, $\left({\lambda }_1\right)$
$\frac{1}{{\lambda }_1}=R(\frac{1}{1^2}-\frac{1}{\infty })$

$\therefore {\lambda }_1=\frac{1}{R}----\left(1\right)$

Similarly, for maximum wavelength , ${\lambda }_2$
$\frac{1}{{\lambda }_2}=R(\frac{1}{1^2}-\frac{1}{2^2})$
$\Rightarrow \frac{1}{{\lambda }_2}=R\left(\frac{3}{4}\right)$
$\therefore {\lambda }_2=\frac{4}{3R}----\left(2\right)$

Dividing equation $\left(1\right)$ and $\left(2\right)$ we get,

$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\frac{1}{R}\times \frac{3R}{4}=\frac{3}{4}$

Hence, $\left(A\right)$ is the correct answer.

Why this Question? This question is based on wavelength of the emitted radiation, during transition of an electron.