A 620 nm light falls on a photoelectric surface and electrons with the maximum kinetic energy of 0.14 eV are emitted. (a) Determine the work function (in eV). eV (b) Determine the cutoff frequency of the surface (in THz). THz (c) What is the stopping potential (in V) when the surface is illuminated with light of wavelength 420 nm

(a) The work function is 1.86 eV.

(b) The cut off frequency is 450 THz.

(c) The stopping potential is 1.16 V.

Explanation:

incident wavelength = 620 nm

Kinetic energy, K = 0.14 eV

According to the photoelectric equation

E = W + KE

where, W is the work function, KE is the kinetic energy.

(a) Let the work function is W.

(b) Let the cut off frequency is f.

W = h f

(c) Let the stopping potential is V.

reserve is the answer

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards