Answer:
(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.
[tex]W = E - KE\\W = \frac{h c}{\lambda }- KE\\W =\frac{6.63\times 10^{-34}\times3\times 10^{8}}{620\times 10^{-9}\times 1.6\times 10^{-19}}-0.14\\\\W =1.86 eV[/tex]
(b) Let the cut off frequency is f.
W = h f
[tex]1.86\times 1.6\times 10^{-19} = 6.63\times 10^{-34}\times f\\f = 4.5\times 10^{14} Hz =450 THz[/tex]
(c) Let the stopping potential is V.
[tex]E = W + eV\\\frac{6.63\times 10^{-34}\times 3\times 10^{8}}{420\times 10^{-9}\times 1.6\times 10^{-19}}=1.8 + eV\\\\V = 1.16 V[/tex]
