Answer
(a) $E_0 = 2.3~eV$
(b) $\lambda_1 = 244~nm$
Work Step by Step
(a) We can write an expression for the maximum kinetic energy:
$K_{max} = \frac{hc}{\lambda} - E_0,~~~$ (where $E_0$ is the work function)
Let $E_1 = \frac{hc}{\lambda_1}$ be the initial energy of each photon of light.
Then: $K_{max} = E_1 - E_0$
If the wavelength is increased by 50%, then the energy of each photon decreases to $\frac{2~E_1}{3}$
We can set up two equations as follows:
(1) $K_{max} = E_1 - E_0 = 2.8~eV$
(1) $E_1 - E_0 = 2.8~eV$
(2) $K_{max} = \frac{2~E_1}{3} - E_0 = 1.1~eV$
(2) $-E_1 + 1.5~E_0 = -1.65~eV$
We can add the two equations to find $E_0$:
$0.5~E_0 = 1.15~eV$
$E_0 = 2.3~eV$
(b) We can find the initial wavelength:
$K_{max} = \frac{hc}{\lambda_1} - E_0 = 2.8~eV$
$\frac{hc}{\lambda_1} = 2.8~eV+E_0$
$\lambda_1 = \frac{hc}{2.8~eV+E_0}$
$\lambda_1 = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(2.8~eV+2.3~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda_1 = 2.44\times 10^{-7}~m$
$\lambda_1 = 244~nm$
