Question

The electric field at a point associated with a light wave is given by
E = 200[sin(6 × 1015)t + sin(9 × 1015)t] Vm-1
Given : h = 4.14 × 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

• A. 1.90 eV
• B. 3.27 eV
• C. 3.60 eV
• D. 3.42 eV

Answer 1

Answer: (A)

1.90 eV

Answer 2

The correct answer is (D) : 3.42 eV
Frequency of EM waves
$= \frac{6}{2π} × 10^{15}$
and
$\frac{9}{2π} × 10^{15}$
Energy of one photon of these waves
$=(4.14 × 10^{-15} × \frac{6}{2π} × 10^{15})eV$
and
$(4.14 × 10^{-15} × \frac{9}{2π} ×10^{15})eV$
3.95 eV and 5.93 eV
⇒ Energy of maximum energetic electrons
= 5.93 – 2.50
= 3.43 eV