Question

The maximum speed of photoelectrons ejected from the surface of a metal is 4.07E5 m/s, when light with a wavelength of 185 nm is shined on the surface. What is the work function of the metal? 2.40 eV 5.17 eV 6.24 eV 4.70 eV 1.42 eV

Answer 1

If h$\nu$ is the energy of the incident photon, is the work function and Emax is the maximum kinetic energy of the electron, then by the equation of photoelectric effect

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$h\nu-\phi_{0}=E_{max}$

Фо - hv Етаг

0o = hv - (Umar)

$\phi_{0}=\frac{hc}{\lambda}-\frac{1}{2}m(v_{max})^{2}$

Given v(max) = 4.07 x 10^5m/s and $\lambda$ = 185nm. Also h = 6.626 x 10^-34Js, c = 3 x 10^8 m/s and m = 9.1 x 10^-31kg. Therefore,

$\phi_{0}=\frac{6.626\times 10^{-34}\times 3\times 10^{8}}{185\times 10^{-9}}-\frac{1}{2}\times 9.\times 10^{-31}\times (4.07\times 10^{5})^{2}$

$\phi_{0}=9.99\times 10^{-19}J$

But,

$1eV = 1.6\times 10^{-19}J$

Therefore,

$\phi_{0} =\frac{9.99\times 10^{-19}}{1.6\times 10^{-19}}=6.24eV$

So the work function of the metal is 6.24eV.

Third option is the correct answer.