Discussion Forum

The ratio of slopes of K_max   vs  V and V₀  vs v curves  in trhe photoelectric effects gives (v= freequency, K_max = maximum kinetic energy ,  v₀ = stopping potential)

Answer:

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+ K_max

or    K_max= hv- hv₀

  Thus, slope of K_max  vs v is 

$(slope)_{2}=h\therefore\frac{(slope)_{2}}{(slope)_{1}}=\frac{h}{h/e}=e$