Question

The surface of a metal of work function 2 eV is illuminated by light whose electric field component is $E=100\left[\sin \left(1.5\times {10}^{15}{s}^{-1}\right)t\right]$. Find the approximate maximum kinetic energy of photoelectrons emitted from the surface.

• A. 2 eV
• B. 4.2 eV
• C. 1 eV
• D. zero

Answer 1

The correct option is B 4.2 eV

Work function $=2eV$

$E=100\left[\sin (1.5\times {10}^{15}{s}^{-1})t\right]$

Now,

Kinetic energy $=\frac{1}{2}{mV}^2$

The light has contained two different frequencies. The one with large frequency will cause photo electrons with largest kinetic energy.

The largest frequency is

$V=\frac{W}{2\pi }=\frac{1.5\times {10}^{15}}{2\times \pi }=2.3\times {10}^{14}{s}^{-1}$

The maximum kinetic energy of photo electrons is $K_{max}=hV-W$

$=(6.626\times {10}^{-34}\times 2.3\times {10}^{+14})-2cV=4.2eV$