Question

Calculate the wavelength of radiation emitted, producing a line in Lyman series, when an electron falls from fourth stationary state in hydrogen atom. $[Given:\left(\frac{1}{R_h}\right)=912\stackrel{o}{A}]$

• A. $972.8\stackrel{o}{A}$
• B. $856.7\stackrel{o}{A}$
• C. $654.5\stackrel{o}{A}$
• D. $798.6\stackrel{o}{A}$

Answer 1

The correct option is A $972.8\stackrel{o}{A}$

For electron transition, $\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})=109700\times 1(\frac{1}{1^2}-\frac{1}{4^2})$ so

$\lambda =972.8\stackrel{o}{A}$