HOMEWORK SOLUTIONS

WEEK 2 PROBLEMS

 

Ch.1

#31.) Show that the momentum of a particle having charge e moving in a circle of radius R in a magnetic field is given by

 

p = 300BR

 

Where p is in MeV/c, B is in Teslas, and R is in meters.

 

This problem is worked out for you in example 1.11, page 34

 

Use eq. 1.38

 

p = qBR

 

   = eBR = 1.602 x 10^-19BR kg m/s

 

1 MeV/c = 5.34 x 10^-22 kg m/s

 

/p = 300BR MeV/c

 

 

 

Ch. 2

#13.) The work function for potassium is 2.24 eV. If potassium metal is illuminated with light of wavelength 350 nm, find

 

(a) the maximum kinetic energy of the photoelectrons

 

The maximum kinetic energy of the photoelectrons is simply what’s given by the photoelectric equation, eq. 2.24

 

K_Max = hf - f

 

We’re given the work function, so all we need to do is convert the given wavelength into a frequency

 

c = lf

 

f = c/l

 

Plug and chug

 

                          K_Max = 1.30 eV

 

and (b) the cutoff wavelength

 

The cutoff wavelength is the wavelength where K_Max = 0.

 

                          K_Max = hc/l - f = 0

 

                          l = hc/f = 554 nm

 

 

#19.) A light source of wavelength l illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1.00 eV. A second light source with half the wavelength of the first ejects photoelectrons with a maximum kinetic energy of 4.00 eV. What is the work function of the metal?

 

Set up the photoelectric equation for the first light with wavelength l

 

                          K_Max1 = hc/l₁ - f             (1)

 

Where K_Max = 1 eV.

 

Now, set up the photoelectric equation for the second light, which has wavelength l₂ = l₁/2

 

                          K_Max2 = hc/l₂ - f

 

                                    = 2hc/l₁ - f          (2)

 

Solve eq. (1) for l₁

 

                          l₁ = hc/(K_Max1 + f)

 

Substitute this expression for l₁ into eq. (2)

 

                          K_Max2 = 2(K_Max1 + f) - f

 

Solve for f

 

                          f = K_Max2 – 2K_Max1

                             = 2.00 eV