http://www.ck12.org Chapter 23. Quantum PhysicsIt is given that the solar power output is 3. 83 × 1026 Js. A photon of wavelength 500 nm has energyE=hcλ=(^6.^626 ×^10− (^34) J−s)( 3. 00 × 108 ms)
500 × 10 −^9 m =^3.^976 ×^10
− (^19) → 3. 98 × 10 − (^19) J.
Therefore, the number of photons is
#o f photons=^3.^83 ×^10
26 Js
3. 976 × 10 −^19 J=^9.^63 ×^10
44
As a comparison, the number of photons emitted each second by a 100-W light bulb is 2. 52 × 1020.
Illustrative Example 24.2.2
a. InFigure23.6, a beam of monochromatic light is incident upon a potassium photocell with work function 2. 2 eV.
If the stopping potential of the ejected electrons is 0.90 V, what is the energy of the incident photons?
Solution:
The work required to bring one electron to rest is
W=eV= ( 1. 00 × 10 −^19 C)( 0. 90 V) = 0. 90 eV
But,
KEmax=W
KEmax= 0. 90 eV→KEmax=h f−Wo→ 0. 90 eV=h f− 2. 2 eV→h f= 3. 1 eV
b. What is the frequency of the light?
Solution:
f=Eh=
( 3. 1 eV)( 1. 60 × 10 −^19 eVJ)
6. 626 × 10 −^34 J−s =^7.^49 ×^10
(^14) Hz→ 7. 5 × 1014 Hz
Light of this frequency is violet.
For more information on the photoelectric effect follow the link below.
http://www.youtube.com/watch?v=RR4T1IlM2Jw&feature=related