Question

Electromagnetic radiation whose electric component varies with time as
$E=C_1(C_2+C_3\cos \omega t)\cos {\omega }_{0}t$
Here $C_1$,$C_2$ and $C_3$ are constants, is incident on lithium and liberates photoelectrons. If the kinetic energy of most energetic electrons be $2.6eV$, the work function of lithium is (in $eV$).

[Take: ${\omega }_0=2.4\pi \times {10}^{15}rad/sec$ and $\omega =8\pi \times {10}^{14}rad/sec$, planks constant $h=6.6\times {10}^{-34}MKS$]

Answer 1

Given, $E=C_1(C_2+C_{3}cos\omega t)cos{\omega }_{0}t$

or $E=C_1{C}_{2}cos{\omega }_{0}t+C_1{C}_{3}cos\omega tcos{\omega }_{0}t$

or $E=C_1^{\prime }cos{\omega }_{0}t+C_2^{\prime }cos({\omega }_0-\omega )t+C_2^{\prime }cos({\omega }_0+\omega )t$ where $C_1^{\prime }=C_1{C}_2,C_2^{\prime }=C_1{C}_3/2$

Hence, the above equation indicates the wave is the superposition three waves of frequencies - ${\omega }_0,({\omega }_0-\omega )$ and $({\omega }_0+\omega )$

So, maximum frequency is ${\omega }_{max}=({\omega }_0+\omega )$

The photoelectric equation , $h{\nu }_{max}=K_{max}+W$

or $W=h\frac{w_o+w}{2\pi }-K_{max}=6.6\times {10}^{-34}\times \frac{(2.4\pi \times {10}^{15})+(8\pi \times {10}^{14})}{2\pi }\times \frac{1}{1.6\times {10}^{-19}}-2.6$ where $(1eV=1.6\times {10}^{-19}J)$

or $W=6.6-2.6=4eV$