Answer to Question #228107 in Atomic and Nuclear Physics for Mis

Question

Calculate the velocity of an electron emitted by radiation of frequency 5.7 x 10^14Hz from a surface with a work function of 3.18 x 10^-19J

Answer 1

The photoelectric equation is

$\frac{hc}{λ} = ϕ+E_K$

E_K=Kinetic energy $=\frac{1}{2}mv^2$

$\frac{hc}{λ} = ϕ+\frac{1}{2}mv^2$

h=plank's constant

c=speed of light

λ=wavelength

ϕ=work function

m=mass of electron

v=velocity of electron

Given

$h=6.626 \times 10^{-34} \;Js \\ c= 3 \times 10^8 \;m/s \\ ϕ = 3.18 \times 10^{-19} \;J \\ v= 5.7 \times 10^{14} \;Hz \\ m= 9.10938 \times 10^{-31} \;kg \\ λ = \frac{c}{frequency} \\ = \frac{3 \times 10^8}{5.7 \times 10^{14}} = 5.26 \times 10^{-7} \;m \\ v = \sqrt{\frac{2}{m} [\frac{hc}{λ }-ϕ] } \\ = \sqrt{\frac{2}{9.10938 \times 10^{-31}} [\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{5.26 \times 10^{-7} }-3.18 \times 10^{-19}] } \\ = \sqrt{2.1955 \times 10^{30} \times (3.779 \times 10^{-19} -3.18 \times 10^{-19} ) } \\ = \sqrt{1.3151 \times 10^{11}} \\ = 3.626 \times 10^5\; m/s$

Answer: $3.626 \times 10^5 \; m/s$