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.
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Conductivity would be the property.
Answer:
.
Explanation:
Since no external force is acting on the system.
Therefore, Total energy remains constant before and after.
So, Total energy of system= energy due to potential applied+kinetic energy
(Here v=velocity ,V=potential ,q=charge and m=mass).
Putting values .
We get, .
At point B charged particle is moving faster as compared to point A.
Hence, it is the required solution.
Answer:
180m to the east
Explanation:
Displacement is the distance traveled in a specific direction. It is a vector quantity with both magnitude and direction. Therefore, the start and finish position is very paramount.
point A runs 150m east,
70m west
100m east
150m
--------------------------------------------------------→
70m
←---------------------
100m
-----------------------------------→
The displacement of the athlete = 150 - 80 + 100 = 180m to the east
Answer:
a) Velocity = 4.2m/s
b) Acceleration = 2.94m/s^2
c) Force exerted on the floor= 1401.4×10^3N
Explanation:
a) Velocity,V=sqrt(2×9.8×0.900)
V= 4.2m/s
b) Vf2= V^2+2ay2
a= 4.2^2 - 0/2×3
a= 17.64/6= 2.94m/s^2
c) Newton's 2nd law indicates:
Fnet= F - mg=ma
F= m(g+a)
F=110(9.8+2.94)
F=110×12.94
F= 1401.4N
