# Solve for wavelength of photon given threshold frequency

The threshold frequency of a potassium metal plate is $$\pu{4.35\times10^14 Hz}$$. Determine the wavelength of a photon required to reach $$\pu{2.50 eV}$$ of the maximum kinetic energy of an emitted electron.

Options:
A. $$\ \ \ \pu{2.88\times10^2 nm}$$
B. $$\pu{-4.96\times10^2 nm}$$
C. $$\ \ \ \pu{9.50\times10^-21 nm}$$
D. $$\ \ \ \pu{6.69\times10^2 nm}$$
E. $$\ \ \ \pu{2.02\times10^15 nm}$$

From what I understand, I used the equation $$KE = h\nu + h\nu_0$$ and replaced $$\nu$$ with $$\frac cA$$ because wavelength and frequency are inversely proportional. Then I plugged in the given components, $$\nu_0$$ and a fraction of $$KE$$ to solve; however, I don't understand how to approach this as I'm only given $$2.5 \pu{eV} (4.005 \times 10^{-19} \pu J)$$ of maximum $$KE$$, not an actual value for $$KE$$. The answer I'm getting is $$\pu{1.14 \times 10^{-12} nm}$$ if I just use $$\pu{2.5 eV} = KE$$ max, which isn't an answer choice, and the given answer is A. $$2.88 \times 10^2 \pu{nm}$$. Please help.

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– A.K.
Sep 22 '18 at 6:10
• Please do not remove formatting markups without cause. They make the question more presentable.
– A.K.
Sep 22 '18 at 19:56

Always work in energy not wavelength. You are given $$\nu_0$$ in Hz ($$\mathrm{s}^{-1}$$), multiply this by $$h$$ to convert it to Joules; ($$h$$ has units J s). You have the kinetic energy in eV, use the conversion $$1\mathrm{eV} = 1.602\cdot 10^{-19}$$ J to get into joules. Now you have only to find energy $$h\nu$$ by subtraction. Finally convert energy to frequency using $$E=h\nu$$ and then to wavelength using $$c=\lambda \nu$$. Make sure $$c$$ is in m/s and as you answer will be in meters then convert to nm from metres.