VITEEE 2015 Paper

hv = hv0+ev0
v0 =

h

e

v−

h

e

v0
On comparing this equation with the straight line equation, i.e y = mx + c The slope of v0 vs v is (v0 is stopping potential)
(slope)1 =

h

e

Likewise,
hv = hv0 + Kmax
or Kmax = hv - hv0
Thus, slope of Kmax vs v is (slope)2 = h
∴

(slope)1

(slope)2

=

h

h∕e

= e