Physics - Capacitor Question with Solution | TestHub

Given: Power of lamp P = 0.10 W Distance r = 1.0 m Effective area A = 4 cm² =

$4.0\times 10^{-4},m^2$

Wavelength λ = 2537 Å = 253.7 nm Work function φ = 2.22 eV hc ≈ 1240 eV·nm 1 eV =

$1.602\times 10^{-19},J$

Photon energy (in eV)

$E_{ph}(eV)=\frac{hc}{\lambda (nm)}=\frac{1240}{253.7}=4.887,eV\approx 4.90,eV$

Photon energy (in J)

$E_{ph}(J)=4.887\times 1.602\times 10^{-19}=7.83\times 10^{-19},J$

Power incident on the photosensitive area Fraction of lamp power hitting the area (assuming isotropic emission):

$f=\frac{A}{4\pi r^2}=\frac{4.0\times 10^{-4}}{4\pi \times 1^2}=\frac{1.0\times 10^{-4}}{\pi }$

Power on cathode:

$P_c=P\times f=0.10\times \left(\frac{1.0\times 10^{-4}}{\pi }\right)=\frac{1.0\times 10^{-5}}{\pi }=3.1831\times 10^{-6},W;(approximately)$

Photon flux (number of photons received per second)

$\dot{N}=\frac{P_c}{E_{ph}(J)}=\frac{3.1831\times 10^{-6}}{7.83\times 10^{-19}}\approx 4.06\times 10^{12},photons/s$

Cut-off (stopping) potential

$V_s=E_{ph}(eV)-\varphi =4.887-2.22=2.667,V\approx 2.68,V$

Answers (photon energy, photon flux, cut-off potential): = 4.90 eV;

$4.06\times 10^{12}$

photons/s; 2.68 V

This corresponds to option (C).