In the photoelectric experiment, if we use a monochromatic light, the I-V curve is as shown. If work function of the metal is 2 eV, estimate the power of light used (in W). (Assume efficiency of photoemission =10-3 %, i.e. number of photoelectrons emitted are 10-3 % of number of photons incident on metal.)

Question

In the photoelectric experiment, if we use a monochromatic light, the I-V curve is as shown. If work function of the metal is 2 eV, estimate the power of light used (in W). (Assume efficiency of photoemission =10-3 %, i.e. number of photoelectrons emitted are 10-3 % of number of photons incident on metal.) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The energy of incident photons is given by h v=e Vs+φ0=2+5=7 eV ( VS is stopping potential and φ0 is work function) ⇒ Saturation current =10-5 A =(η P/h v) e=(10-5 P/7 × e) e Thus P=7 W

The energy of incident photons is given by hv=eVs​+ϕ0​=2+5=7eV ( VS​ is stopping potential and ϕ0​ is work function) ⇒ Saturation current =10−5A =hvηP​e=7×e10−5P​e Thus P=7W

The energy of incident photons is given by
$h v = e V_{s} + ϕ_{0} = 2 + 5 = 7 e V$
( $V_{S}$ is stopping potential and $ϕ_{0}$ is work function)
$\Rightarrow$ Saturation current $= 1 0^{- 5} A$
$= \frac{η P}{h v} e = \frac{1 0 ^{- 5} P}{7 \times e} e$
Thus $P = 7 W$