## Chemistry 9th Edition

$\lambda=9.38\times 10^{-8}m=93.8nm$
We can find the longest wavelength from $n=6$ to $n=5$ as $\lambda=\frac{hc}{(-R_H)(\frac{1}{n_f^2}-\frac{1}{n_i^2})}$ We plug in the known values to obtain: $\lambda=\frac{6.63\times 10^{-34}(3\times 10^8)}{-2.18\times 10^{-18}(\frac{1}{5^2}-\frac{1}{6^2})}$ $\lambda=7.46\times 10^{-6}m=7460nm$ Now we can find the longest wavelength from $n=6$ to $n=1$ as $\lambda=\frac{hc}{(-R_H)(\frac{1}{n_f^2}-\frac{1}{n_i^2})}$ We plug in the known values to obtain: $\lambda=\frac{6.63\times 10^{-34}(3\times 10^8)}{-2.18\times 10^{-18}(\frac{1}{1^2}-\frac{1}{6^2})}$ $\lambda=9.38\times 10^{-8}m=93.8nm$