Problem on photoelectric effect

[issacnewton]

Hello

Here is the problem I am trying to solve.
Find the maximum kinetic energy of the photo electron liberated from the surface of lithium
by electromagnetic radiation whose electric component varies with time as

[tex]E=a(1+\cos \omega t)\;\cos(\omega_0 t)[/tex]

where a is constant and [itex]\omega = 6 \times 10^{14} \; s^{-1}[/itex] and
[itex]\omega_0 = 3.6 \times 10^{15} \; s^{-1}[/itex]

options for the answers are given as

a)0 eV
b)2 eV
c) 0.38 eV
d) 0.5 eV

Now since maximum kinetic energy will depend upon the maximum frequency present in the
E. So I was trying to turn the equation for E in the following way using trig identities.

[tex]E=a\left[\cos \omega_0 t + \frac{1}{2}\cos(\omega_0 - \omega)t+\frac{1}{2}\cos(\omega_0 + \omega)t \right][/tex]

And now I can see that the max frequency component present is [itex]\omega_0 + \omega[/itex]
But [itex]\omega_0 + \omega = 2\pi(f_0 +f)[/itex]. Now using the photoelectric equation and
noting that the work function for lithium (wikipedia) is 2.3 eV, we have

[tex]K_{\mathrm{max}} = h\;(f_0 +f) - \phi[/tex]

Now since [itex]\phi[/itex] is given in eV, we can use value of h in eV-sec, so we get

[tex]K_{\mathrm{max}} = (4.14\times 10^{-15})\left[\frac{6\times 10^{14} + 3.6\times 10^{15}}{2\pi}\right]-2.3[/tex]

and rounding numbers , I get approximately 0.5 eV, which is option d). Does it look ok ?

thanks

 

[ehild]

Yes, it looks OK if the work function is correct. .

ehild

 

[issacnewton]

thanks ehild