Question

Electromagnetic radiations described by the equation: $E=100[\sin (2\times {10}^{15})t+\sin (6.28\times {10}^{15})t]V/m$ are incident on a photosensitive surface having work function $1.5eV$. The maximum kinetic energy of photoelectrons will be

• A. $0.125eV$
• B. $2.625eV$
• C. $4.125eV$
• D. $6.125eV$

Answer 1

The correct option is B $2.625eV$

Standard form will be like $E=E_0\left[sin\right({\omega }_{1}t)+sin({\omega }_{2}t\left)\right]$

so we have ${\omega }_1=2\times {10}^{15}rad/s$ and ${\omega }_2=6.28\times {10}^{15}rad/s$

corresponding frequency will be ${\nu }_1={\omega }_1/2\pi =\frac{2\times {10}^{15}}{6.28}$

and ${\nu }_2={\omega }_2/2\pi =\frac{6.28\times {10}^{15}}{6.28}={10}^{15}Hz$

We can see that ${\nu }_2$ is $higher$ so it will provide $maximum$ kinetic energy.

Energy of the ${\nu }_2$ photon will be $E=h{\nu }_2=6.6\times {10}^{-34}Js\times {10}^{15}Hz=6.6\times {10}^{-19}Joule=\frac{6.6\times {10}^{-19}Joule}{1.6\times {10}^{-19}J/eV}$

or $E=4.125eV$

Consumed energy in work function will be $W=1.5eV$

So remaining energy that is maximum kinetic energy will be $KE=E-W=(4.125-1.5)eV=2.625eV$

Correct option Is B.