In an experiment on photoelectric effect, a student plots stopping potential `V_(0)` against reciprocal of the wavelength `lamda` of the incident light for two different metals A and B. These are shown in the figure

Looking at the graphs, you can most appropriately say that :

Options

A. `V_2 lt V_1`
B. `V_1 lt V_(2) lt 2V_1`
C. `V_2 = 2V_1`
D. `V_2 gt 2V_1`

Answer

D

Explanation

According to Einstein.s photoelectric equation
`K_(max) = hupsilon - phi_(0)`
`eV_(s)= (hc)/(lamda) - phi_(0) implies V_(s) =(hc)/(lamda_e)-(phi_0)/e`
where, `lamda` = Wavelength of incident light
`phi_0` = Work function
`V_s` = Stopping potential
According to given problem
`V_1=(hc)/(lamdae)-(phi_0)/(e)" "...(i) and V_(2) (hc)/((lamda/2)e) -(phi_0)/e " "...(ii)`
`V_2 = (2hc)/(lamdae)-(phi_0)/e=(2hc)/(lamdae) -(2phi_0)/(e) +(2phi_0)/(e)-(phi_0)/e =2 ((hc)/(lamdae) -(phi_e)/e)+(phi_0)/e`
`V_(2)=2V_(1) + phi_(0) :. V_(2) gt 2V_(1)" " ` (Using (i))