In which of the following transitions will the wavelength of light emitted be minimum?

n = 2 to n = 1

The correct answer is Option (4) - n = 2 to n = 1

$\frac{1}{λ}=R\left(\frac{1}{{n_f}^2}-\frac{1}{{n_i}^2}\right)$

$n_i=5,n_f=4$

$\frac{1}{λ}=R\left(\frac{1}{16}-\frac{1}{25}\right)=R(0.0625-0.04)=0.0225R$

$n_i=2,n_f=1$

$\frac{1}{λ}=R\left(1-\frac{1}{4}\right)=\frac{R3}{4}=0.75R$

$n_i=4,n_f=3$

$\frac{1}{λ}=R\left(\frac{1}{9}-\frac{1}{16}\right)=R(0.11-0.0625)=0.575R$

$n_i=3,n_f=2$

$\frac{1}{λ}=R\left(\frac{1}{4}-\frac{1}{9}\right)=0.14R$

∴ n = 2 to n = 1 has the highest value of $\left(\frac{1}{{n_f}^2}-\frac{1}{{n_i}^2}\right)$ and thereby will have least λ.