Answer
When $n = 6$:
$\lambda = 410.06~nm$
When $n = 8$:
$\lambda = 388.80~nm$
When $n = 10$:
$\lambda = 379.69~nm$
Work Step by Step
We can use the following equation to find the wavelengths of spectral lines in the Balmer series:
$\frac{1}{\lambda} = R~(\frac{1}{2^2}- \frac{1}{n^2})$
Note that $R$ is the Rydberg constant and $R = 1.0974\times 10^7~m^{-1}$
When $n = 6$:
$\frac{1}{\lambda} = R~(\frac{1}{2^2}- \frac{1}{n^2})$
$\frac{1}{\lambda} = (1.0974\times 10^7~m^{-1})~(\frac{1}{2^2}- \frac{1}{6^2})$
$\frac{1}{\lambda} = (1.0974\times 10^7~m^{-1})~(\frac{2}{9})$
$\frac{1}{\lambda} = 0.24387\times 10^7~m^{-1}$
$\lambda = 4.1006\times 10^{-7}~m$
$\lambda = 410.06~nm$
When $n = 8$:
$\frac{1}{\lambda} = R~(\frac{1}{2^2}- \frac{1}{n^2})$
$\frac{1}{\lambda} = (1.0974\times 10^7~m^{-1})~(\frac{1}{2^2}- \frac{1}{8^2})$
$\frac{1}{\lambda} = (1.0974\times 10^7~m^{-1})~(\frac{15}{64})$
$\frac{1}{\lambda} = 0.2572\times 10^7~m^{-1}$
$\lambda = 3.8880\times 10^{-7}~m$
$\lambda = 388.80~nm$
When $n = 10$:
$\frac{1}{\lambda} = R~(\frac{1}{2^2}- \frac{1}{n^2})$
$\frac{1}{\lambda} = (1.0974\times 10^7~m^{-1})~(\frac{1}{2^2}- \frac{1}{10^2})$
$\frac{1}{\lambda} = (1.0974\times 10^7~m^{-1})~(\frac{6}{25})$
$\frac{1}{\lambda} = 0.263376\times 10^7~m^{-1}$
$\lambda = 3.7969\times 10^{-7}~m$
$\lambda = 379.69~nm$
