The work functions of Ag, Mg, K and Na respectively in eV are 4.3, 3.7, 2.25, 2.30. when an electromagnetic radiation of wavelength of 300 nm is allowed to fall on these metal surfaces, the number of metals from which the electrons are ejected is. `(1 eV = 1.6022 times 10^-19 J)`

Options

A. 4
B. 3
C. 2
D. 5

Answer

B

Explanation

Work function for Ag. Mg. K and Na in Joules are
for Ag `= 4 . 3 eV xx 1.6022 xx 10^(-19) = 6 . 89 xx 10^(-19) J`
` Mg = 3 . 7 eV xx 1.6 022 xx 10^(-19) = 5 . 93 xx 10^(-19) J`
`K =2 . 25 eV xx 1.6022 xx 10^(-19) = 3 . 60 xx 10^(-19) J `
` Na = 2 . 30 eV + 1.6022 xx 10^(-19) = 3 . 68 xx 10 ^(-19) J`
Energy used is of wave length `(lambda = 300 nm)`
`E_(300) = (hc)/(lambda) = (6.6 xx 10^(-34) xx 3 xx 10^(8))/(300 xx 10^(-9))`
`E_(300) = (19.80 xx 10 ^(-26))/(300 xx 10^(-9)) = 0.066 xx 10^(-17)`
`= 6.6 xx 10^(-19) J`
`:.` Metal will eject electrons only when `W le E_(300)`
`:.` Only Mg. K and Na (3-metals) will eject the electrons.