A light source of wavelength λ illuminates a metal and ejects photo-electrons with (K.E)max=1eV. Another light source of wavelength λ/3 ejects photo-electrons from the same metal with (K.E)max=4eV. Find the value of Work Function?

`phi = hnu_0 = "0.50 eV"`

As seen in the photoelectric effect, the kinetic energy `K` is leftover from the incoming light source energy `hnu` that is in excess of the threshold energy `hnu_0`:

`K = hnu - hnu_0`

where `nu_0` is the threshold frequency, and `hnu_0` is otherwise known as the "work function".

From here we set up a system of equations to describe what is given in the question.

`"1 eV" = color(white)(.)(hc)/lambda - hnu_0``" "" "" "bb((1))`

`"4 eV" = (3hc)/(lambda) - hnu_0``" "" "" "bb((2))`

where `nu = c//lambda` for light waves and `lambda` is its wavelength. `c = 2.998 xx 10^8 "m/s"` is the speed of light.

Here we take `-3cdot(1) + (2)` to get:

`" "-"3 eV" = -(3hc)/lambda + 3hnu_0`
`+ul(" ""4 eV" = " "(3hc)/(lambda) - color(white)(.)hnu_0)`
`" "color(white)(/)"1 eV" = " "" "" "" "2hnu_0`

So, the threshold energy is:

`color(blue)(phi = hnu_0 = "0.50 eV")`

As a check, does it work? From `(1)`:

`"1 eV" stackrel(?" ")(=) (hc)/lambda - "0.50 eV"`

`=> (hc)/lambda = "1.50 eV"`

Plugging this into `(2)`:

`"4 eV" stackrel(?" ")(=) (3hc)/lambda - "0.50 eV"`

`stackrel(?" ")(=) 3 ("1.50 eV") - "0.50 eV"`

`=` `"4 eV"` `color(blue)(sqrt"")`