The electric field at a point associated with a light wave is given by
E = 200 [sin (6 $\times$ 1015)t + sin (9 $\times$ 1015)t] Vm$-$1
Given : h = 4.14 $\times$ 10$-$15 eVs
If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be
Solution
<p>Frequency of EM waves = ${6 \over {2\pi }} \times {10^{15}}$ and ${9 \over {2\pi }} \times {10^{15}}$</p>
<p>Energy of one photon of these waves</p>
<p>$$ = \left( {4.14 \times {{10}^{ - 15}} \times {6 \over {2\pi }} \times {{10}^{15}}} \right)$$ eV</p>
<p>and $$\left( {4.14 \times {{10}^{ - 15}} \times {9 \over {2\pi }} \times {{10}^{15}}} \right)$$ eV</p>
<p>= 3.95 eV and 5.93 eV</p>
<p>$\Rightarrow$ Energy of maximum energetic electrons</p>
<p>= 5.93 $-$ 2.50 = 3.43 eV</p>
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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