Question

Calculate the wavelength of radiation emitted producing a line in Lyman series, when an electron falls from fourth stationary state in hydrogen atom $\left(R_{H}=1 \cdot 1 \times 10^{7} m ^{-1}\right)$.

Answer 1

We know that $\frac{1}{\lambda}=R_{H}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

In Lyman series, $n_{1}=1$. Here $n_{2}=4$

$\therefore \quad \frac{1}{\lambda}=1 \cdot 1 \times 10^{7}\left(1-\frac{1}{16}\right)=1 \cdot 1 \times 10^{7} \times \frac{15}{16}$

$\therefore \quad \lambda=\frac{16}{15 \times 1 \cdot 1} \times 10^{-7} m =9.7 \times 10^{-8} m .$