TS EAMCET 9 May 2024 Shift 1 Solved Paper

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