Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V. The work function of the used metal in the experiment is $$\alpha \times 10^{-19}$$ J. The value of $$\alpha$$ is _______. (h = $$6.62 \times 10^{-34}$$ J s, e = $$1.6 \times 10^{-19}$$ C and c = $$3 \times 10^8$$ m/s)

Use photoelectric equation:

work function = hν − eV₀

step 1: find photon energy

$$E=\frac{hc}{\lambda}$$

$$=\frac{6.62\times10^{-34}\times3\times10^8}{331\times10^{-9}}$$

$$J=\frac{1.986\times10^{-25}}{3.31\times10^{-7}}\approx6\times10^{-19}$$

step 2: subtract stopping energy

$$eV_0=1.6\times10^{-19}\times0.2=0.32\times10^{-19}$$

step 3: work function

$$\phi=6\times10^{-19}-0.32\times10^{-19}=5.68\times10^{-19}$$