The electric field at a point associated with a light wave is given by

E = 200 $\left[sin\left(6\times 10^{15}\right)t+sin\left(9\times 10^{15}\right)t\right]V{m}^{-1}$

Given : h = 4.14 × 10^-15 eVs

If the light fall on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be

<p>The light wave contains two lights of different frequencies, so</p><p><math><mrow><msub><mrow><mi>E</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>=</mo><mi>h</mi><msub><mrow><mi>υ</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>=</mo><mfrac><mrow><mn>4</mn><mo>.</mo><mn>1</mn><mn>4</mn><mo>×</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mo>−</mo><mn>1</mn><mn>5</mn></mrow></msup><mo>×</mo><mn>6</mn><mo>×</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>1</mn><mn>5</mn></mrow></msup></mrow><mrow><mn>2</mn><mi>π</mi></mrow></mfrac><mo>=</mo><mn>3</mn><mo>.</mo><mn>9</mn><mn>6</mn><mi>e</mi><mi>V</mi><mo>, </mo><mtext> </mtext><mtext> </mtext><mi>a</mi><mi>n</mi><mi>d</mi></mrow></math></p><p><math><mrow><msub><mrow><mi>E</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>=</mo><mi>h</mi><msub><mrow><mi>υ</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>=</mo><mfrac><mrow><mn>4</mn><mo>.</mo><mn>1</mn><mn>4</mn><mo>×</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mo>−</mo><mn>1</mn><mn>5</mn></mrow></msup><mo>×</mo><mn>9</mn><mo>×</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>1</mn><mn>5</mn></mrow></msup></mrow><mrow><mn>2</mn><mi>π</mi></mrow></mfrac><mo>=</mo><mn>5</mn><mo>.</mo><mn>9</mn><mn>2</mn><mi>e</mi><mi>V</mi></mrow></math></p><p>Maximum kinetic energy of the photoelectron = 5.9-2.5 = 3.42 eV</p>

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