Answer 1

The slope of the cut-off voltage (V) versus frequency (ν) of an incident light is given as:

`V/v = 4.12 xx 10^(-15) Vs`

V is related to frequency by the equation:

`hv = eV`

Where

e = Charge on an electron = 1.6 × 10⁻¹⁹ C

h = Planck’s constant

`:. h = e xx V/v`

`= 1.6 xx 10^(-19) xx 4.12 xx 10^(-15) = 6.592 xx 10^(-34) Js`

Therefore, the value of Planck’s constant is `6.592 xx 10^(-34) Js`