Question

A certain metal of work function 1.6 eV is irradiated with an ultra violent light of wave length 3.6*10⁴ m. Calculate the maximum

Kinetic energy of ejected electrons in joules

And speed of an emitted electrons

Answer 1

a) We can find the maximum kinetic energy of the photoelectrons emitted from the formula:

$$KE_{max}=hf-W,$$ $$KE_{max}=\dfrac{hc}{\lambda}-W,$$ $$KE_{max}=\dfrac{6.63\cdot10^{-34}\ J\cdot s\cdot3\cdot10^8\ \dfrac{m}{s}}{360\cdot10^{-9}\ m}-1.6\ eV,$$ $$KE_{max}=5.52\cdot10^{-19}\ J-1.6\ eV\cdot \dfrac{1.6\cdot10^{-19}\ J}{1\ eV}=2.96\cdot10^{-19}\ J.$$

b) We can find the speed of an emitted electrons from the definition of the kinetic energy:

$$KE=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{\dfrac{2KE}{m}},$$ $$v=\sqrt{\dfrac{2\cdot 2.96\cdot10^{-19}\ J}{9.1\cdot10^{-31}\ kg}}=8.06\cdot10^5\ \dfrac{m}{s}.$$

Answer:

a) $KE_{max}=2.96\cdot10^{-19}\ J.$

b) $v=8.06\cdot10^5\ \dfrac{m}{s}.$