Question

23.   Photons and Energy  

   A. A certain source emits radiation of wavelength 500.0 nm. Determine its frequency. Calculate the energy associated with this photon at 500.0 nm.

   B. If it takes 3.36 x 10-19 J of energy to eject an electron from the surface of a certain metal, Calculate the frequency of this energy and the longest possible wavelength in nm.
  
   C. Ionization energy is the energy required to remove an electron from an atom in the gas phase. The ionization energy of gold is 890.1 kJ/mol. Calculate the minimum wavelength of light that will ionize gold.

Answer 1

    A]

Convert nm to m:

500.0 nm = 500.0 x 10¯9 m = 5.000 x 10¯7 m

Determine the frequency:

1) λν = c

(5.000 x 10¯7 m) (x) = 3.00 x 108 m/s

x = 6.00 x 1014 s¯1

2) Determine the energy:

E = hν:

x = (6.626 x 10¯34 J s) (6.00 x 1014 s¯1)

x = 3.9756 x 10¯19 J

B]

E = hν

3.36 x 10-19 J = (6.626 x 10¯34 J s) (x)

x = 5.071 x 1014 s¯1

2) Determine the longest possible wavelength:

λν = c

(x) (5.071 x 1014 s¯1) = 3.00 x 108 m/s

x = 5.916 x 10¯7 m

Wavelength = 592 nm

C]

E / atom = (890.1 x 10E03) / (6.02 x 10E23)

= 1.479 x 10^-18 J / atom

E = (1.48 x 10^-18)

= (6.64 x 10^-34 x 3 x 10⁸) / (1.48 x 10^-18)

= 134.6 x 10^-9 m

= 134.6 nm