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

Answer

4

Explanation

The energy associated with incident photon
`= (hc)/(lambda)n implies E = (6.6xx10^(-34) x 10^(8))/(300xx0^(-9))`
`E` in eV `=(6.6xx10^(-34) xx 3 xx 10^8)/( 300xx10^(-9) xx 1.6xx 10^(-19)) = 4.16 eV`
So , number of metals showing photo-electric effects will be (4) i.e., Li, Na , K , Mg