question_answer

Photoelectric work function of a metal is 1 eV and light of wavelength \[\lambda \] = \[\text{3000 }\overset{\text{o}}{\mathop{\text{A}}}\,\] falls on it. The photoelectrons come out with velocity

A)  \[\text{1}{{0}^{\text{2}}}\text{ m}/\text{s}\]

B)  \[\text{1}{{0}^{\text{4}}}\text{ m}/\text{s}\]

C)  \[\text{1}{{0}^{\text{6}}}\text{ Ws}\]  

D)  \[\text{1}{{0}^{\text{8}}}\text{ m}/\text{s}\]