Given that a photon of light of wavelength 10,000 mathringA has an energy equal to 1.23 eV. When light of wavelength 5000 mathringA and intensity I0 falls on a photoelectric cell, the saturation current is 0.40 × 10-6 A and the stopping potential is 1.36 V; then the work function is

Question

Given that a photon of light of wavelength 10,000 mathringA has an energy equal to 1.23 eV. When light of wavelength 5000 mathringA and intensity I0 falls on a photoelectric cell, the saturation current is 0.40 × 10-6 A and the stopping potential is 1.36 V; then the work function is (A) 0.43 eV (B) 1.10 eV (C) 1.36 eV (D) 2.47 eV

Tardigrade Admin · Accepted Answer

Since E propto (1/λ), so energy corresponding to 5000 mathringA is E=2.46 eV Now, h v-W=e Vs or 2.46 eV -W=1.36 eV W=(2.46-1.36) eV =1.1 eV

Q. Given that a photon of light of wavelength 10,000A˚ has an energy equal to 1.23eV. When light of wavelength 5000A˚ and intensity I0​ falls on a photoelectric cell, the saturation current is 0.40×10−6A and the stopping potential is 1.36V; then the work function is

0.43 eV

1.10 eV

1.36 eV

2.47 eV

Since E∝λ1​, so energy corresponding to 5000A˚ is E=2.46eV Now, hv−W=eVs​ or 2.46eV−W=1.36eV W=(2.46−1.36)eV=1.1eV

Since $E \propto \frac{1}{λ}$ , so energy corresponding to $5000 \overset{˚}{A}$ is
$E = 2.46 e V$
Now, $h v - W = e V_{s}$
or $2.46 e V - W = 1.36 e V$
$W = (2.46 - 1.36) e V = 1.1 e V$