Radiation of different frequencies fall on two different metal surfaces. The . . .

Question : Radiation of different frequencies fall on two different metal surfaces. The graph below represents the kinetic energy of the emitted photoelectrons as a function of frequencies of incident radiations.

  

a) If the two metals along with their work functions are : Potassium (2.3 eV) and Aluminum (4.3 eV), identify which line represents each metal. Give a reason for your answer. 

b) Write the photoelectric equation that represents each of the above graphs. Find the slope of the graph. 

Doubt by Tushar

Solution : 

a) We know, 

According to Einstein Photoelectric Equation

K_max=hν-hν₀

OR

K_max=hν-Ф₀

For a constant frequency of incident radiation (ν ), K_max is more for Metal 1 than Metal 2.

K_max=hν  -Ф₀

From the above equation it is clear than K_max is higher for lower Ф₀.

Hence, Metal 1 must represent Potassium of lower work function 2.3 eV while the Metal 2 must represent Aluminum of higher work function 4.3 eV.

b) 

According to Einstein Photoelectric Equation

K_max=hν-hν₀

K_max=hν+(-hν₀)

on comparing the above equation with y=mx+c

we get, 
the slope of the K_max vs 'ν' graph 
Slope = h
Y-intercept of K_max vs 'ν' graph 

Y-intercept = -hν₀

Photoelectric equation for Metal 1

K_max=hν₁-hν₀₁

Photoelectric equation for Metal 2

K_max=hν₂-hν₀₂