Question

The surface of Lithium of work function ᶲ is illuminated by electromagnetic radiation whose electric field component varies with time as a=(1+Cos(omega)t)Cos(omega0)t . The maximum kinetic energy of photo electron liberated from surface is?

Answer 1

Answer on Question #64028-Physics-Atomic and Nuclear Physics

The surface of Lithium of work function $\Phi$ is illuminated by electromagnetic radiation whose electric field component varies with time as $a = (1 + \cos(\omega \omega a)t)\cos(\omega \omega a)t$ . The maximum kinetic energy of photo electron liberated from surface is?

Solution

$$a = (1 + \cos (\omega) t) \cos (\omega_ {0}) t = \cos (\omega_ {0}) t + \cos (\omega) t \cos (\omega_ {0}) t$$

Using trigonometric identity:

$$\cos (\omega) t \cos (\omega_ {0}) = \frac {1}{2} (\cos (\omega + \omega_ {0}) t + \cos (\omega - \omega_ {0}) t).$$

Thus,

$$a = \cos (\omega_ {0}) t + \frac {1}{2} \cos (\omega + \omega_ {0}) t + \frac {1}{2} \cos (\omega - \omega_ {0}) t)$$

The maximum angular frequency is $(\omega + \omega_0)$ .

The maximum frequency of electromagnetic radiation is

$$f = \frac {(\omega + \omega_ {0})}{2 \pi}.$$

The maximum kinetic energy of photo electron liberated from surface is

$$K = h \frac {(\omega + \omega_ {0})}{2 \pi} - \phi = \hbar (\omega + \omega_ {0}) - \phi .$$

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