Discussion Forum



1)

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

 


A) the ratio of Planck's constant of electronic charge

B) work function

C) Planck's constant

D) charge of electron

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 

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