CHAPTER 16. OPTICAL PHENOMENA; PROPERTIES OF MATTER 16.3
UV radiation wavelength = 250 nm = 250 × 10 −^9 m
Planck’s constant: h = 6, 63 × 10 −^34 m^2 kgs−^1
speed of light: c = 3× 108 ms−^1Step 2 : Solve the problemEk =
hc
λ
− φ= [6, 63 × 10 −^34 ×
3 × 108
250 × 10 −^9
]− 6 , 9 × 10 −^19
= 1, 06 × 10 −^19 J
The maximum kinetic energy of the emitted electron will be
1 , 06 × 10 −^19 J.Example 2: The photoelectric effect - II
QUESTIONIf we were to shine thesame ultraviolet radiation (f = 1, 2 × 1015 Hz), on a
gold foil (work function = 8, 2 × 10 −^19 J), would any electrons be emitted from
the surface of the gold foil?SOLUTIONFor the electrons to beemitted from the surface, the energy of each photon
needs to be greater than the work functionof the material.Step 1 : Calculate the energy of the incident photonsEphoton = hf
= 6, 63 × 10 −^34 × 1 , 2 × 1015
= 7, 96 × 10 −^19 JTherefore each photonof ultraviolet light has an energy of
7 , 96 × 10 −^19 J.Step 2 : Write down the work function for gold.φgold = 8, 2 × 10 −^19 J