Answer:
Option D
Explanation:
hv= hv0 +ev0
$v_{0}=\frac{h}{e}v-\frac{h}{e}v_{0}$
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)
$(slope)_{1}=\frac{h}{e}$
Like wise,
hv= hv0+ Kmax
or Kmax= hv- hv0
Thus, slope of Kmax vs v is
$(slope)_{2}=h\therefore\frac{(slope)_{2}}{(slope)_{1}}=\frac{h}{h/e}=e$