QJEEM22Ph2a3c: JEE Main 2022 Physics - Class 12 JEEM22Ph2a3c Solution: The electric field at a point associated with a light wave is…

Question

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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

Accepted Answer

Frequency of EM waves = ${6 \over {2\pi }} \times {10^{15}}$ and ${9 \over {2\pi }} \times {10^{15}}$

Energy of one photon of these waves

$= \left( {4.14 \times {{10}^{ - 15}} \times {6 \over {2\pi }} \times {{10}^{15}}} \right)$ eV

and $\left( {4.14 \times {{10}^{ - 15}} \times {9 \over {2\pi }} \times {{10}^{15}}} \right)$ eV

= 3.95 eV and 5.93 eV

$\Rightarrow$ Energy of maximum energetic electrons

= 5.93 $-$ 2.50 = 3.43 eV

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