Answer:
B. blocks 2 & 3.
Explanation:
Block 1 has equal & opposite forces acting on it.
Block 2 has 5N on one side, 3N on the other. It will move in the direction the 5N of force is pushing.
Block 3 has no opposing force.
If the light of wavelength 700 nm strikes such a photocathode the maximum kinetic energy, in eV, of the emitted electrons is 0.558 eV.
so - $KE_{max} = hc/lembda} work
threshold when KE = 0
hc/lambda = work = 1240/900=1.38 eV
b) Kemax = hc/lambda - work = 1240/640 -1.38=0.558 eV
What is photocathode?
- A photocathode electrolyte interface can be used in a photoelectrolysis cell as the primary light-harvesting junction (in conjunction with an appropriate electrochemical anode) or as an optically complementary photoactive half-cell in a tandem photoelectrode photoelectrolysis cell (Hamnett, 1982; Kocha et al, 1994).
- In the case of the former, the electrode should ideally harvest photon energy across the majority of the solar spectrum in order to achieve the highest energy conversion efficiency possible.
- In the latter case, however, the photocathode may only be active in a specific band of the solar spectrum in order to generate a cathodic photocurrent sufficient to match the current generated in the photoanodic half-cell.
To learn more about Photocathode from the given link:
brainly.com/question/9861585
#SPJ4
Answer:
24445.85 J/s
Explanation:
Area, A = 300 m^2
T = 33° C = 33 + 273 = 306 k
To = 18° C = 18 + 273 = 291 k
emissivity, e = 0.9
Use the Stefan's Boltzman law
Where, e be the energy radiated per unit time, σ be the Stefan's constant, e be the emissivity, T be the temperature of the body and To be the absolute temperature of surroundings.
The value of Stefan's constant, σ = 5.67 x 10^-8 W/m^2k^4
By substituting the values
E = 24445.85 J/s
