Prove that the product of the slope of `V_(0)`-f graph and electronic charge gives the value of planck's constant.

Explanation

From Einstein's photoelectric equation we know that,
`(1)/(2) mv_("max")^(2) = hf - W_(0), " But " (1)/(2) mv_("max")^(2) = eV_(0)`
`:. eV_(0) = hf - W_(0)`
or, `V_(0) = (h)/(e) f - (W_(0))/(e)`...(1)
So the plot of stopping potential `(V_(0))` versus frequency (f) will be a straight line.
Comparing equation (1) with the equation of a straight line,
`y = mx +c`, we get the slope of `V_(0)-f` graph, `m = (h)/(e)`
`:.` Planck's constant, h = me