The work function `(W_(0))` of Li, K, Mg, Ag and Cu are 2.42, 2.25, 3.70, 4.80 eV respectively. The number of metals which undergo photoelectric effect if a radiation of wavelength 540 nm falls on them is
(1 eV = `1.602xx10^(-19)` J)

Options

A. 4
B. 2
C. 1
D. 3

Answer

C

Explanation

For any metal to show photoelectric effect, `v gt v_(0)` is threshold frequency, v is the frequency of light.
Given,
`W_(0)` of `Li = 2.42 eV =2.42 xx 1.602 xx 10^(-19) J`
`= 3.876 xx 10^(-19) J`
`W_(0)` of `K = 2.25 eV = 3.604 xx 10^(-19) J`
`W_(0)` of `Mg = 3.70 eV = 5.927 xx 10^(-19) J`
`W_(0)` of `Ag = 4.30 eV = 6.888 xx 10^(-19) J`
`W_(0)` of `Cu = 4.80 eV = 7.689 xx 10^(-19)J`
Also `W_(0) = hv_(0)`
`because v_(0)` of `Li = W_(0)//h`
`= (3.876 xx 10^(-19))/(6.626 xx 10^(-34)) = 0.584 xx 10^(15) s^(-1)`
`:. v_(0)` of `K = (W_(0))/(h) = (3.604 xx 10^(-19))/(6.626 xx 10^(-34))`
`= 0.54 xx 10^(15) s^(-1)`
`:. v_(0)` of `Mg = (5.927 xx 10^(-19))/(6.626 xx 10^(-34)) = 0.89 xx 10^(15) s^(-1)`
`:. v_(0)` of `Ag = (6.888 xx 10^(-19))/(6.626 xx 10^(-34)) = 1.03 xx 10^(15) s^(-1)`
`v_(0)` of `Cu = (7.689 xx 10^(-19))/(6.626 xx 10^(-34)) = 1.16 xx 10^(15) s^(-1)`
Frequency of light can be calculated as :
`v = (c )/(lambda)`
`= (3 xx 10^(8))/(540 xx 10^(-9)) = 0.0056 xx 10^(17) s^(-1)`
`~~0.56 xx 10^(15) s^(-1)`
Thus, only in case of K, `v gt v_(0)` thus it will show photoelectric effect.