• question_answer Given that a photon of light of wavelength 10,000 angstrom has an energy equal to 1.23 eV. When light of wavelength 5000 angstrom and intensity${{I}_{0}}$ falls on a photoelectric cell, the saturation current is$0.40\times {{10}^{-6}}$ampere and the stopping potential is volt; then the work function is: A)  0.43 eV                      B)  1.10 eVC)  1.36 eV                      D)  2.47 eV

$E=hv=\frac{hc}{\lambda }$or $\frac{E}{E'}=\frac{\lambda '}{\lambda }$ or $\frac{E}{1.23}=\frac{10,000}{5,000}$or $E=2.46\,eV$ Now, $hv-W=\frac{1}{2}m{{({{v}_{\max }})}^{2}}$or $hv-W=e{{V}_{s}}$ $2.46\,eV-W=e(1.36)$volt or $W=(2.46-1.36)=1.1\,eV$ Hence, the correction option is .