The maximum energies of photoelectrons emitted by a metal are `E_(1) and E_(2)` when the incident radiation has frequencies `f_(1) and f_(2)` respectively. Show that Planck's constant h and the work function `W_(0)` of the metal are
`h = (E_(1) - E_(2))/(f_(1) - f_(2)), W_(0) = (E_(1) f_(2) - E_(2) f_(1))/(f_(1) - f_(2))`

Explanation

According to Einstein's photoelectric equation we have, `E_(1) = hf_(1) - W_(0)`…(1)
`E_(2) = hf_(2) - W_(0)`…(2)
`E_(1) - E_(2) = h(f_(1) - f_(2))`
`:. h= (E_(1) - E_(2))/(f_(1) - f_(2))` [for `E_(1) gt E_(2)`] ….(3)
Again multiplying equation (1) by `f_(2)` and equation (2) by `f_(1)` we obtain,
`E_(1) f_(2) = hf_(1) f_(2) - f_(2) W_(0)`...(4)
and `E_(2) f_(1) = hf_(1) f_(2) - f_(1) W_(0)`....(5)
Subtracting equation (5) from equation (4), we get
`E_(1) f_(2) - E_(2) f_(1) = f_(1) W_(0) - f_(2) W_(0) = W_(0) (f_(1) - f_(2))`
`:. W_(0) = (E_(1) f_(2) - E_(2) f_(1))/(f_(1) - f_(2))`