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

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

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 ........... $eV$

  • A
    $1.90$
  • B
    $3.27$
  • C
    $3.60$
  • $3.42$

Answer

Correct option: D.
$3.42$
d
For maximum $KE$ we will take higher frequency $\left(f=\frac{9 \times 10^{15}}{2 \pi} Hz \right)$

$K_{\max }=h f-\phi$

$=\frac{9 \times 10^{15} \times 4.14 \times 10^{-15}}{2 \pi}-2.50$

$3.43 \,eV$

nearest is $3.42 \,eV$

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