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  • The maximum kinetic energy of photo electron liberated from the surface of lithium <img alt="(\phi=2.39 \mathrm{eV})" src="https://entrancecorner.oncodecogs.com/gif.latex?%28%5Cphi%3D2.39%20%5

The maximum kinetic energy of photo electron liberated from the surface of lithium (\phi=2.39 \mathrm{eV})  by EM wave whose electric component varies with time as  E=a(1+\cos \omega t) \cdot \cos \omega_0 t  is:
(a. is a constant, \omega=6 \times 10^{14} \mathrm{rad} / \mathrm{s}
\left.\omega_0=3.6 \times 10^{15} \mathrm{rad} / \mathrm{s}\right)
 

Option: 1

0.29 \mathrm{ev}


Option: 2

0.32 \mathrm{ev}


Option: 3

 0.34 \mathrm{ev}


Option: 4

0.37 \mathrm{ev}


Answers (1)

best_answer


E=a(1+\cos \omega t) \cos \omega_0 t

\begin{aligned} & \Rightarrow E=a \cos \omega_0 t+a \cos \omega t \cdot \cos \omega_0 t \\ & \Rightarrow E=a \cos \omega_0 t+\frac{1}{2} \cdot a \cos \left(\omega+\omega_0\right) t+\frac{1}{2} \cos \left(\omega-\omega_0\right) t \end{aligned}
This is a complex vibration consisting of harmonic components of frequencies \omega_0, \omega+\omega_0  and  \omega-\omega_0

The highest angular frequency is \omega+\omega_0

$ Now $K \cdot E_{\max }=h \nu-\phi=h \frac{\left(\omega+\omega_0\right)}{2 \pi}-\phi

 K \cdot E_{\max }=\frac{6.6 \times 10^{-34}}{2 \pi} .\left[\begin{array}{ll} \left. 6 \times 10^{14}+3.6 \times 10^{15}\right]-2.39 \times 1.6 \times 10^{-19}\end{array}\right.

 

\begin{aligned} K \cdot E_{\text {max }} & =4.41 \times 10^{-19}-3.82 \times 10^{-19} \mathrm{~J} \\ & =0.59 \times 10^{-19} \mathrm{~J}=0.37 \mathrm{eV} \end{aligned}

Posted by

avinash.dongre

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