Answer
$\lambda_{41} = 97.4~ nm$
$\lambda_{42} = 487~nm$
$\lambda_{43} = 1.88~\mu m$
Work Step by Step
We can find the energy when $n = 1$:
$E_1 = -\frac{13.6~eV}{1^2} = -13.6~eV$
We can find the energy when $n = 2$:
$E_2 = -\frac{13.6~eV}{2^2} = -3.4~eV$
We can find the energy when $n = 3$:
$E_3 = -\frac{13.6~eV}{3^2} = -1.51~eV$
We can find the energy when $n = 4$:
$E_4 = -\frac{13.6~eV}{4^2} = -0.85~eV$
One wavelength is associated with the jump from $n = 4$ to $n = 1$.
In this case, the energy difference is: $~~\Delta E = (-0.85~eV)-(-13.6~eV) = 12.75~eV$
We can find the wavelength $\lambda_{41}$:
$E = \frac{h~c}{\lambda}$
$\lambda_{41} = \frac{h~c}{E}$
$\lambda_{41} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(12.75~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda _{41}= 9.74\times 10^{-8}~m$
$\lambda_{41} = 97.4~ nm$
One wavelength is associated with the jump from $n = 4$ to $n = 2$.
In this case, the energy difference is: $~~\Delta E = (-0.85~eV)-(-3.4~eV) = 2.55~eV$
We can find the wavelength $\lambda_{42}$:
$E = \frac{h~c}{\lambda}$
$\lambda_{42} = \frac{h~c}{E}$
$\lambda_{42} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(2.55~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda _{42}= 4.87\times 10^{-7}~m$
$\lambda_{42} = 487~nm$
One wavelength is associated with the jump from $n = 4$ to $n = 3$.
In this case, the energy difference is: $~~\Delta E = (-0.85~eV)-(-1.51~eV) = 0.66~eV$
We can find the wavelength $\lambda_{43}$:
$E = \frac{h~c}{\lambda}$
$\lambda_{43} = \frac{h~c}{E}$
$\lambda_{43} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(0.66~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda _{43}= 1.88\times 10^{-6}~m$
$\lambda_{43} = 1.88~\mu m$
