Given that
Caesium metal work function = 2.14 eV
Incident light frequency \upsilon = 6*10^{14} Hz.
a) Maximum kinetic energy of the emitted electrons
Photon's energy = energy needed for the emission of an electron (work function) + Maximum kinetic energy of the electron
E = W + K.E
K.E = h \upsilon - W
E = \frac{ 6.63 * 10^{-34} * 6*10^{14}}{1.6*10^{-19}} - 2.14 eV
K.E = 2.485 - 2.140
K.E = 0.345 eV
Hence, Maximum kinetic energy of the emitted electrons K.E = 0.345 eV
b) Stopping potential
By conservation of energy, kinetic energy has to be equal to the change in potential energy.
K.E = eV
V = \frac{K.E}{e}
V = \frac{0.345 eV}{1.6*10^{-19}} = \frac{0.345 * 1.6*10^{-19} V}{1.6*10^{-19}}
V = 0.345 V
c) Maximum speed of the emitted photoelectrons?
Mass of the electron m = 9.1*10^{-31}
Kinetic energy K.E = \frac{1}{2} mv^2
v^2 = \frac{2K.E}{m}
v^2 = \frac{2*0.345 eV}{9.1*10^{-31}}
v^2 = \frac{2*0.345 * 6.1*10^{-19}}{9.1*10^{-31}}
v = \sqrt{\frac{2*0.345 * 6.1*10^{-19}}{9.1*10^{-31}}}
v = 3.323 * 10^{5} m/s
v = 332.3 km/s
Hence, Maximum speed of the emitted photoelectrons v = 332.3 km/s