Answer
When $n = 2$:
$\lambda = 121.50~nm$
When $n = 3$:
$\lambda = 102.52~nm$
When $n = 4$:
$\lambda = 97.20~nm$
When $n = 5$:
$\lambda = 94.92~nm$
Work Step by Step
To calculate the wavelengths of spectral lines in the Lyman series, we can use the following equation:
$\frac{1}{\lambda} = R~(\frac{1}{1^2}- \frac{1}{n^2})$
Note that $R$ is the Rydberg constant and $R = 1.0974\times 10^7~m^{-1}$
When $n = 2$:
$\frac{1}{\lambda} = R~(\frac{1}{1^2}- \frac{1}{n^2})$
$\frac{1}{\lambda} = (1.0974\times 10^7~m^{-1})~(\frac{1}{1^2}- \frac{1}{2^2})$
$\frac{1}{\lambda} = (1.0974\times 10^7~m^{-1})~(\frac{3}{4})$
$\frac{1}{\lambda} = 0.82305\times 10^7~m^{-1}$
$\lambda = 1.2150\times 10^{-7}~m$
$\lambda = 121.50~nm$
When $n = 3$:
$\frac{1}{\lambda} = R~(\frac{1}{1^2}- \frac{1}{n^2})$
$\frac{1}{\lambda} = (1.0974\times 10^7~m^{-1})~(\frac{1}{1^2}- \frac{1}{3^2})$
$\frac{1}{\lambda} = (1.0974\times 10^7~m^{-1})~(\frac{8}{9})$
$\frac{1}{\lambda} = 0.975467\times 10^7~m^{-1}$
$\lambda = 1.0252\times 10^{-7}~m$
$\lambda = 102.52~nm$
When $n = 4$:
$\frac{1}{\lambda} = R~(\frac{1}{1^2}- \frac{1}{n^2})$
$\frac{1}{\lambda} = (1.0974\times 10^7~m^{-1})~(\frac{1}{1^2}- \frac{1}{4^2})$
$\frac{1}{\lambda} = (1.0974\times 10^7~m^{-1})~(\frac{15}{16})$
$\frac{1}{\lambda} = 1.0288125\times 10^7~m^{-1}$
$\lambda = 0.9720\times 10^{-7}~m$
$\lambda = 97.20~nm$
When $n = 5$:
$\frac{1}{\lambda} = R~(\frac{1}{1^2}- \frac{1}{n^2})$
$\frac{1}{\lambda} = (1.0974\times 10^7~m^{-1})~(\frac{1}{1^2}- \frac{1}{5^2})$
$\frac{1}{\lambda} = (1.0974\times 10^7~m^{-1})~(\frac{24}{25})$
$\frac{1}{\lambda} = 1.053504\times 10^7~m^{-1}$
$\lambda = 0.9492\times 10^{-7}~m$
$\lambda = 94.92~nm$
