Question 63: Dual Nature of Matter and Radiation

Question

A light wave described by $E=60\left[\sin \left(3 	imes 10^{15}ight) t+\sin \left(12 	imes 10^{15}ight) tight]$ (in SI units) falls on a metal surface of work function 2.8 eV . The maximum kinetic energy of ejected photoelectron is (approximately)
$\_\_\_\_$ eV. $\left(h=6.6 	imes 10^{-34}ight.$ J.s. and $\left.e=1.6 	imes 10^{-19} \mathrm{C}ight)$

Accepted Answer

According to Einstein's photoelectric effect, the energy of an incident photon (E) on a metal of work function $(\phi)$ then the maximum kinetic energy $\left(\mathrm{K}_{\max }ight)$ to the ejected electron is,
$$ \mathrm{K}_{\max }=\mathrm{E}-\phi $$
The energy of a photon is given by :
$$ \mathrm{E}=\mathrm{hf}=\frac{\mathrm{h} \omega}{2 \pi} $$
Where :
h is Planck's constant.
$\omega$ is the angular frequency of the light wave.
The given electric field is $\mathrm{E}=60\left[\sin \left(3 	imes 10^{15}ight) \mathrm{t}+\sin \left(12 	imes 10^{15}ight) \mathrm{t}ight]$
The given electric field equation contains two sinusoidal terms with angular frequencies are $\omega_1=3 	imes 10^{15} \mathrm{rad} / \mathrm{s}$ and $\omega_2=12 	imes 10^{15} \mathrm{rad} / \mathrm{s}$
Since $\mathrm{K}_{\max }$ depends on the maximum photon energy, we must use the higher angular frequency, $\omega_{\max }=12 	imes 10^{15} \mathrm{rad} / \mathrm{s}$, because higher frequency corresponds to higher energy per photon.
The energy in Joules is $E=\frac{h \omega}{2 \pi}$.
$$ \mathrm{E}(	ext { in } \mathrm{eV})=\frac{\mathrm{h} \cdot \omega_{\max }}{2 \pi \cdot \mathrm{e}} $$
Substituting the given values :
$$ E=\frac{\left(6.6 	imes 10^{-34}ight) 	imes\left(12 	imes 10^{15}ight)}{2 	imes 3.14 	imes\left(1.6 	imes 10^{-19}ight)} $$
$$\Rightarrow $$ $E=\frac{79.2 	imes 10^{-19}}{10.048 	imes 10^{-19}}$
$$\Rightarrow $$ $$ \mathrm{E} \approx 7.88 \mathrm{eV} $$
So, the maximum kinetic energy of ejected electron is,
$$ \mathrm{K}_{\max }=\mathrm{E}-\phi $$
$$\Rightarrow $$ $\mathrm{K}_{\max }=7.88 \mathrm{eV}-2.8 \mathrm{eV}$
$$\Rightarrow $$ $\mathrm{K}_{\max }=5.08 \mathrm{eV}$
$$\Rightarrow $$ $$ \mathrm{K}_{\max } \approx 5.1 \mathrm{eV} $$
Therefore, the kinetic energy of electron is approximately 5.1 eV.
Hence, the correct option is (D).

Answer

3.8

Answer

7.8

Answer

6.0

Answer

5.1

Dual Nature of Matter and Radiation

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