All electrons ejected from a metallic surface by incident light of wavelength 400 nm travelled 1 m in the direction of uniform electric field of `2 NC^(-1)` and came to rest. The work function of the surface is

Options

A. `1.1 eV `
B. `2.2 eV`
C. `3.1 eV`
D. `5.1 eV`

Answer

A

Explanation

Given wavelength of incident radiation
`lambda =400 nm =4 xx 10^(-7) m`
electric field E =2 `N//C` and distance s=1 m
Energy of the incident light
`E = (hc )/( lambda) = (6.6 xx 10^(34) xx 3xx 10^(8))/( 4xx 10^(-7))`
`=4.95 xx 10^(-19) J = (4.95 xx 10^(19))/( 1.6 xx 10^(-19))`
`J = 3.09 = 3l eV`
If a be the retardation of emitted electrons in the electric field .
Initial speed u of emitted electron is calculated as
`v^(2) = u^(2) 2as rArr 0 = u^(2) - 2a xx l rArr u^(2) =2a = (2qE )/( m )`
`:.` Maximum kinetic energy of the electron
`K_(max) = 32 xx 10^(19) J = 2eV`
`:.` Work function of the surface
`W_(0) = E - E_(max) = 3.1 -2 =1.1 eV`