Question

When light of wavelength, 115 nm, impinges a photocathode that has a binding energy of 2.59 eV, electrons are ejected. What is the speed (in m/s) of the ejected electrons? h = 6.626 x 10–34J·s, c = 2.9979 x 108m/s, 1 eV = 1.602 x 10–19 J, NA= 6.022 x 1023mol–1, me = 9.11 x 10–31 kg, KE = ½ mv2. Report your answer to three significant figures. It might be easier to enter your answer in scientific notation. So, for example, if your answer is 923672, in scientific notation and to 3 significant figures, the number is 9.24 x 105and you would enter 9.24e5.

Answer 1

Given that

binding energy (or) work function W = 2.59 eV = 2.59 x 1.602 x 10^–19 J = 4.15 x 10^–19 J

Given that

wavelength λ = 115 nm = 115 x 10^-9 m ( 1 nm = 10^-9 m )

Energy E = hc / λ

h = planck's constant = 6.625 x 10^-34 J.s

c = velocity of light = 3 x 10⁸ m/s

Hence,

  E = hc / λ

=  (6.625 x 10^-34 J.s) (3 x 10⁸ m/s) / (115 x 10^-9 m)

= 0.173 x 10^-17 J

E =   0.173 x 10^-17 J

We know that E = W + KE KE = kinetic energy

E = W + KE

KE = E - W =  0.173 x 10^-17 J - 4.15 x 10^–19 J = 0.1315 x 10^-17 J

KE = (1/2) mv² m = mass of electron, v = speed of electron

v = [ 2 KE/m ]^1/2

      = [ 2 x 0.1315 x 10^-17 J/ 9.11 x 10^–31 kg ]^1/2

     = 0.17 x 10⁷ m/s

v = 0.17 x 10⁷ m/s

Therefore,

speed of the ejected electrons = 0.17 x 10⁷ m/s