Question

Ratio of wavelength for second line of Balmer series and first line of Lymen series in $H$ -atom is -

• A. $5:27$
• B. $27:5$
• C. $1:4$
• D. $4:1$

Answer 1

Answer: (D)

$\frac{1}{{\left(\lambda \right)}_1}=R{\left(1\right)}^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\dots \left(1\right);{\left(\lambda \right)}_1=\frac{16}{3R}$
$\frac{1}{{\left(\lambda \right)}_2}=R{\left(1\right)}^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\dots \left(2\right);{\left(\lambda \right)}_2=\frac{4}{3R}$
$\frac{{\lambda }_1}{{\lambda }_2}=\frac{16}{4}=\frac{4}{1}$