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QJEEM19Ph1c56: JEE Main 2019 Physics - Class 12 JEEM19Ph1c56 Solution: The electric field of light wave is given as This light falls…

The electric field of light wave is given as </mo></mrow><mover><mi>E</mi><mo>→</mo></mover><mo>=</mo><mrowdata−mjx−texclass="ORD"><msup><mn>10</mn><mrowdata−mjx−texclass="ORD"><mo>−</mo><mn>3</mn></mrow></msup></mrow><mi>cos</mi><modata−mjx−texclass="NONE">⁡</mo><mrowdata−mjx−texclass="INNER"><modata−mjx−texclass="OPEN">(</mo><mrowdata−mjx−texclass="ORD"><mrowdata−mjx−texclass="ORD"><mfrac><mrowdata−mjx−texclass="ORD"><mn>2</mn><mi>π</mi><mi>x</mi></mrow><mrowdata−mjx−texclass="ORD"><mn>5</mn><mo>×</mo><mrowdata−mjx−texclass="ORD"><msup><mrowdata−mjx−texclass="ORD"><mn>10</mn></mrow><mrowdata−mjx−texclass="ORD"><mo>−</mo><mn>7</mn></mrow></msup></mrow></mrow></mfrac></mrow><mo>−</mo><mn>2</mn><mi>π</mi><mo>×</mo><mn>6</mn><mo>×</mo><mrowdata−mjx−texclass="ORD"><msup><mrowdata−mjx−texclass="ORD"><mn>10</mn></mrow><mrowdata−mjx−texclass="ORD"><mn>14</mn></mrow></msup></mrow><mi>t</mi></mrow><modata−mjx−texclass="CLOSE">)</mo></mrow><mover><mrowdata−mjx−texclass="OP"><mi>x</mi></mrow><mo>∧</mo></mover><modata−mjx−texclass="NONE">⁡</mo><mrowdata−mjx−texclass="ORD"><mfrac><mrowdata−mjx−texclass="ORD"><mimathvariant="normal">N</mi></mrow><mi>C</mi></mfrac></mrow></math></mjx−assistive−mml></mjx−container></mjx−assistive−mml></mjx−container></mo></mrow><mover><mi>E</mi><mo>→</mo></mover><mo>=</mo><mrow data-mjx-texclass="ORD"><msup><mn>10</mn><mrow data-mjx-texclass="ORD"><mo>−</mo><mn>3</mn></mrow></msup></mrow><mi>cos</mi><mo data-mjx-texclass="NONE">⁡</mo><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mrow data-mjx-texclass="ORD"><mrow data-mjx-texclass="ORD"><mfrac><mrow data-mjx-texclass="ORD"><mn>2</mn><mi>π</mi><mi>x</mi></mrow><mrow data-mjx-texclass="ORD"><mn>5</mn><mo>×</mo><mrow data-mjx-texclass="ORD"><msup><mrow data-mjx-texclass="ORD"><mn>10</mn></mrow><mrow data-mjx-texclass="ORD"><mo>−</mo><mn>7</mn></mrow></msup></mrow></mrow></mfrac></mrow><mo>−</mo><mn>2</mn><mi>π</mi><mo>×</mo><mn>6</mn><mo>×</mo><mrow data-mjx-texclass="ORD"><msup><mrow data-mjx-texclass="ORD"><mn>10</mn></mrow><mrow data-mjx-texclass="ORD"><mn>14</mn></mrow></msup></mrow><mi>t</mi></mrow><mo data-mjx-texclass="CLOSE">)</mo></mrow><mover><mrow data-mjx-texclass="OP"><mi>x</mi></mrow><mo>∧</mo></mover><mo data-mjx-texclass="NONE">⁡</mo><mrow data-mjx-texclass="ORD"><mfrac><mrow data-mjx-texclass="ORD"><mi mathvariant="normal">N</mi></mrow><mi>C</mi></mfrac></mrow></math></mjx-assistive-mml></mjx-container></mjx-assistive-mml></mjx-container> This light falls on a metal plate of work function 2eV. The stopping potential of the photoelectrons is :
Given, E (in eV) = 12375/λ(inÅ)

Multiple Choice Options

  • A 2.48 V
  • B 0.48 V
  • C 0.72 V
  • D 2.0 V

Step-by-Step Verified Solution

ω=6×1014×2π
f = 6 × 1014
C = f λ

λ=Cf=3×1086×1014=5000 Å
Energy of photon ⇒123755000=2.475eV

From Einstein’s equation
KEmax = E – ϕ
eVs = E – ϕ
eVs = 2.475 – 2
eVo = 0.475 eV
Vo = 0.48 V

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