Suppose the wavelength of the incident light in photoelectric effect experiment is increased from 3000Ao to 3040Ao. Find the corresponding change in the stopping potential. [Take the product hc=12.4×10−7Vm]
ΔV=ΔEincidente
(hcλ1−hcλ2)
=4.133−4.078
=5.5×10−2V