When a high frequency electromagnetic radiation is incident on a metallic surface, electrons are emitted from the surface. Energy of emitted photoelectrons depends only on the frequency of incident electromagnetic radiation and the number of emitted electrons depends only on the intensity of incident light.
Einstein's protoelectron equation `[K_(max)=hv-phi}` correctly ecplains the PE, where `upsilon=` frequency of incident light and `phi=` work function.
Q. Light of wavlength 3300 is incident on two metals A and B, whose work function are 4 eV and 2 eV, respectively. Then

Options

A. A will emit photoelectrons but B will not
B. B will emit photoelectrons but A will not
C. both A and B will not emit photoelectrons
D. neither A nor B will emit photoelectrons

Answer

B

Explanation

The energy of the incident photon is
`E=(hc)/(lamda)=(6.6xx10^(-34)xx3xx10^(8))/(3300xx10^(-10))`
`=3.75` eV
A will not emit photoelectrons because energy of incident photon is less than work function of A.