The relationship between the cut-off voltage V0 and the frequency f of incident light in a photoelectric experiment is given by the equation: eV0=hf−φ where: e is the elementary charge ( 1.6×10−19C ), h is Planck's constant ( 6.63×10−34Js ), φ is the work function of the material. Rearranging the equation provides: V0=‌
h
e
f−‌
φ
e
The slope m of the graph of V0 versus f is given by the factor of f, which is ‌
h
e
. Substituting the given values: m=‌
h
e
=‌
6.63×10−34Js
1.6×10−19C
Calculating this gives: m=‌
6.63
1.6
×10−15≈4.14375×10−15Vs Thus, the correct option is: Option A: 414×10−15Vs
