VISUAL  PHYSICS  ONLINE

7.1   THE NATURE OF LIGHT

        ELECTROMAGNEITC WAVES

 

P71 008

(A)

An FM radio station transmits at a frequency of 102.8 MHz.

What is the energy, in joules and in eV of each photon emitted by the transmitter?

What is the wavelength of the transmitted signal.

 

(B)

The minimum amount of energy needed to eject an electron from a clean aluminium surface is 8.72 × 10–19 J. What is the work function for aluminium in eV? What is the maximum wavelength of incident light that can be shone on this aluminium surface to eject electrons? What is the threshold frequency? What part of the electromagnetic spectrum is the incident radiation?

 

 

 

 

 

 

 

    

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

View solution below only after you have completed the answering the question.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution

(A)

f = 102.8 MHz = 102.8´106 Hz

E = ?  J = ?  eV     l = ? m    

qe = 1.602´10-19 C      c = 3.0´ m.s-1      h = 6.626´10-34 J.s     

E = h f    E(eV) = E(J)/qe      c = f l     l = c / f

E = 6.8´10-26  J = 4.3´10-7  eV    

                      very small photon energy since in radio part of em spectrum

l = 2.9  m

                      long wavelength because of small frequencies

 

(B)  

work function W = 8.72´10-19 J = (8.72´10-19) / (1.602´10-19) eV = 5.44 eV

min energy of photon to produce photo-electrons

          h = 6.626´10-34 J.s       c = 3.00´108 m.s-1

          h f = h c / l = W

          f = W / h = 1.32´1015 Hz

          l = h c / W = 2.28´10-7 m         UV