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Light of wavelength 4000 Å falls on a ph...

Light of wavelength `4000 Å` falls on a photosensitive metal and a negative `2 V` potential stops the emitted electrons. The work function of the material ( in eV) is approximately `( h = 6.6 xx 10^(-34) Js , e = 1.6 xx 10^(-19) C , c = 3 xx 10^(8) ms^(-1))`

A

`1.1`

B

`2.0`

C

`2.2`

D

`3.1`

Text Solution

Verified by Experts

The correct Answer is:
A
Energy (E) (eV) `= (12375)/(4000) = 3.09eV`
`K_("max") = E-phi_0`
`implies 2eV = 3.09 -phi_0`
`implies phi_0=1.09 ~~1.1eV`
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