The stopping potential are `V_(1)` and `V_(2)`. Calculate the `(V_(1)-V_(2))`, if the `lambda_(1)` and `lambda_(2)` are wavelength of incident lights, respectively.

If lambda_v, lambda_x and lambda_m represents the wavelength of visible light X-ray and microwave respectively then :

From the above figure the values of stopping potentials for M_(1) and M_(2) for a frequency v_(3)( gt v_(02)) of the incident radiatioins are V_(1) and V_(2) respectively. Then the slope of the line is equal to

The potential energy of a particle of mass m is given by V(x) = lambda_(1) and lambda_(2) are the de - Brogle wavelength of the particle, when = lex le 1 and xgt 1 repectively ,if the total energy of particle is 2 E_(0) find lambda_(1) // lambda_(2)

A surface receives light of wavelength lambda_1=450 nm, causing the ejection of photo-electrons for which the stopping potential is V_(S_1)=0.2 V. If the radiations of wavelength lambda_2=120 nm are now incident on the surface, the threshold frequency for the surface is

The stopping potential for a certain photocell is 1.2 V when light of wavelength lambda_(1)=400 nm is used. It is 0.6 V, when light of wavelength lambda_(2)=500 nm is used. What would be the stopping potential, when both lambda_(1) and lambda_(2) are incident simultaneously ?