The work function `(phi)` of some metals is listed below . The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal is :

Answer

4

Explanation

Energy associated with the incident photon `= (hc)/(lamda)`
i.e., `E = ((6.6 xx 10^(-34) Js) (3 xx 10^(8) ms^(-1)))/((300 xx 19^(-9) m))`
`= 6.6 xx 10^(-19) J = (6.6 xx 10^(-19))/(1.6 xx 10^(-19)) eV = 4.12 eV`
`:.` Metals showing photoelectric effect will be Li, Na, K and Mg only i.e., 4 metals, (which have work function less than 4.12 eV)