The electric field at the 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 :

Question

The electric field at the 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

Tardigrade Admin · Accepted Answer

For maximum KE we will take higher frequency ( f =(9 × 1015/2 π) Hz ) K max = hf -φ =(9 × 1015 × 4.14 × 10-15/2 π)-2.50 3.43 eV text nearest is 3.42 eV

Q. The electric field at the point associated with a light wave is given by E=200[sin(6×1015)t+sin(9×1015)t]Vm−1 Given : h=4.14×10−15eVs If this light falls on a metal surface having a work function of 2.50eV, the maximum kinetic energy of the photoelectrons will be :

1.90 eV

3.27 eV

3.60 eV

3.42 eV

For maximum KE we will take higher frequency (f=2π9×1015​Hz) Kmax​=hf−ϕ =2π9×1015×4.14×10−15​−2.50 3.43eV nearest is 3.42eV

For maximum KE we will take higher frequency
$(f = \frac{9 \times 1 0 ^{15}}{2 π} Hz)$
$K_{m a x} = h f - ϕ$
$= \frac{9 \times 1 0 ^{15} \times 4.14 \times 1 0 ^{- 15}}{2 π} - 2.50$
$3.43 e V nearest is 3.42 e V$