Answer
(a) $\lambda_{31} = 310~ nm$
$\lambda_{32} = 500~ nm$
$\lambda_{21} = 830~ nm$
(b) $\lambda_{13} = 310~ nm$
$\lambda_{12} = 830~ nm$
Work Step by Step
(a) One wavelength is associated with the jump from $n = 3$ to $n = 1$.
In this case, the energy difference is: $~~\Delta E = 4.0~eV-0.0~eV = 4.0~eV$
We can find the wavelength:
$E = \frac{h~c}{\lambda}$
$\lambda_{31} = \frac{h~c}{E}$
$\lambda_{31} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(4.0~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda _{31}= 3.1\times 10^{-7}~m$
$\lambda_{31} = 310~ nm$
One wavelength is associated with the jump from $n = 3$ to $n = 2$.
In this case, the energy difference is: $~~\Delta E = 4.0~eV-1.5~eV = 2.5~eV$
We can find the wavelength:
$E = \frac{h~c}{\lambda}$
$\lambda_{32} = \frac{h~c}{E}$
$\lambda_{32} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(2.5~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda_{32} = 5.0\times 10^{-7}~m$
$\lambda_{32} = 500~ nm$
One wavelength is associated with the jump from $n = 2$ to $n = 1$.
In this case, the energy difference is: $~~\Delta E = 1.5~eV-0.0~eV = 1.5~eV$
We can find the wavelength:
$E = \frac{h~c}{\lambda}$
$\lambda_{21} = \frac{h~c}{E}$
$\lambda_{21} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(1.5~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda_{21} = 8.3\times 10^{-7}~m$
$\lambda_{21} = 830~ nm$
(b) We can find the wavelength $\lambda_{13}$:
$E = \frac{h~c}{\lambda}$
$\lambda_{13} = \frac{h~c}{E}$
$\lambda_{13} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(4.0~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda _{13}= 3.1\times 10^{-7}~m$
$\lambda_{13} = 310~ nm$
We can find the wavelength $\lambda_{12}$:
$E = \frac{h~c}{\lambda}$
$\lambda_{12} = \frac{h~c}{E}$
$\lambda_{12} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(1.5~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda_{12} = 8.3\times 10^{-7}~m$
$\lambda_{12} = 830~ nm$