The work function of tungsten is 2.8 eV. (1) Calculate the speed of the fastest

Question

The work function of tungsten is 2.8 eV. (1) Calculate the speed of the fastest electrons ejected from a tungsten surface when light whose photon energy is 5.8 eV shines on the surface. (2) Find the wavelength of the photon. (3) Find the linear momentum of the photon A photon undergoes Compton scattering off a stationary free electron. The photon scatters at 80 degree from its initial direction; its initial wavelength is 3.0 times 10-12 m. (1) What is the Compton shift? (2) What is the energy of the outgoing photon? (3) What is the electron's kinetic energy?

Explanation / Answer

A) Work function = 2.8 eV

Energy of photon = 5.8 eV

For the fastest electron, all the excess energy gets converted inti its kinetic energy.

So, for the fastest electron, kinetic energy = excess energy of photon = 5.8-2.8 = 3 eV

1 eV = 1.6*10-19 J, So, total kinetic energy = 3*1.6*10-19 J

Now, kinetic energy of electron= 1/2*mass of electron*v2

Since mass of electron = 9.1*10-31 kg

So , 1/2*9.1*10-31*v2 = 3*1.6*10-19

Thus, v = 1.027*106 m/s

Energy of photon, E = h*c/lambda, where lambda is the wavelength

'h' is planck's constant = 6.626*10-34 J.s

c = speed of light = 3*108 m/s

E = energy = 5.8 eV = 5.8 * 1.6 * 10-19 J

Putting values, we get : lambda = 2.14*10-7 m

Linear momentum, p = E/c = h/lambda

So, we get : p = (5.8*1.6*10-19)/(3*108) = 3.09*10-27 kg.m/s

B)

Initial wavelength, lambda = 3*10-12 m

Let final wavelength be lambda'

Angle of scattering, theta = 80'

We know, lambda'-lambda = (h/mc)*(1-cos theta)

where h = planck's constant, m is mass of electron, c is speed of light

So, putting values we get : lambda' = 5*10-12 m

So, compton shift = increase in wavelength = lambda'-lambda = (5-3)*10-12 = 2*10-12 m

Energy of outgoing photon, E = h*c/lambda' = 6.626*10-34*3*108/5*10-12 = 3.97*10-14 J

Kinetic energy of electron = energy lost by photon in this collision

Energy lost = initial energy - final energy = (h*c/lambda)-(h*c/lambda')

Putting values, we get : Kinetic energy of electron = 2.65*10-14 J

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.