Problems on Blackbody Radiation, Light Quantization and Photoelectric Effect

(Serway et al, pp. 92-93) Answers (R. Egerton)

2. Wien’s displacement law: (lambda)max (T) = 2.898E-3 m . K gives

(lambda)max = 2.898E-3 / (273+35) = 9.41E-6 m = 9.41 micrometers

4. (a) Stefan’s law with epsilon = 1 gives:

power density at the filament surface = (5.67E-8)(3000)^4 = 4.59E6 W/m^2

2. surface area = (power) / (power density) = (75)/(4.59E-6) = 1.63E-5 m^2

10. Energy/sec = 3.74E26 J/s = (photons/sec)(energy per photon)

Photon energy = hf = hc/(lambda) = (6.63E-34)(3.00E8)/(500E-9) = 3.97E-19 J

Photons/sec = (3.74E26)/(3.97E-19) = 9.41E44

12. For cutoff , (work function) = h f = hc/(lambda)

= (6.63E-34)(3.00E8)/(250E-9) = 7.946E-19 J = 4.97 eV

16. Photon energy = hf = hc / (lambda) = 6.62E-19 J = 4.13 eV

Mercury would not liberate photoelectrons, since its work function is higher.

For lithium, (K)max = hf – (phi) = 4.13 – 2.3 = 1.83 eV

For beryllium, (K)max = 4.13 – 3.9 = 0.23 eV

20. (a) Vs = 0 for f = 4.1E14 Hz approximately, corresponding to

0 = (K)max = hf – phi, giving phi = (6.63E-34)(4.1E14) = 2.72E-19 J = 1.7 eV

(b) line on graph is given by (e)(Vs) = hf – (phi),

so h/e = slope = (3.0) / (11.2-4.1)(1E14) = 4.2 E-15

(c) cutoff wavelength = (3.0E8) / (4.1E14) = 7.3 E-7 m = 0.73 micrometers

4. From tables, h/e = (6.626E-34)/(1.602E-19) = 4.136 E-15,