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If light of wavelength 700 nm strikes such a photocathode, what will be the maximum kinetic energy, in eV , of the emitted elect

Oksana_A [137]

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

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3 0

2 years ago

20.0 Ω resistor and a 2.50 μF capacitor are connected in parallel with signal generator. The signal generator produces a sinusoi

Free_Kalibri [48]

Answer:

0.15A

Explanation:

The parameters given are;

R=20.0 Ω

C= 2.50 μF

V= 3.00 V

f= 2.48×10^-3 Hz

Xc= 1/2πFc

Xc= 1/2×3.142 × 2.48×10^-3 × 2.5 ×10^-6

Xc= 25666824.1

Z= 1/√(1/R)^2 +(1/Xc)^2

Z= 1/√[(1/20)^2 +(1/25666824.1)^2]

Z= 1/√(2.5×10^-3) + (1.5×10^-15)

Z= 20 Ω

But

V=IZ

Where;

V= voltage

I= current

Z= impedance

I= V/Z

I= 3.00/20

I= 0.15A

6 0

3 years ago

A person walks from her home at position O to position F by taking a path that is comprised of three displacement vectors travel

lidiya [134]

Answer:

Explanation:

A = 155 m at 25° North of west

B = 92 m due north

C = 64.7 m at an angle 50° East of north

Write the displacements in the vector form

$\overrightarrow{A} = 155\left ( Cos25\widehat{i}-Sin25\widehat{j} \right )$

$\overrightarrow{A}=140.5\widehat{i}-65.5\widehat{j}$

$\overrightarrow{B}=92\widehat{j}$

$\overrightarrow{C} = 64.7\left ( Cos50\widehat{i}+Sin50\widehat{j} \right )$

$\overrightarrow{C} = 41.6\widehat{i}+49.6\widehat{j}$

(a)

$\overrightarrow{A}.\overrightarrow{C}=\left ( 140.5\widehat{i}-65.5\widehat{j} \right ).\left ( 41.6\widehat{i}+49.5\widehat{j} \right )$

$\overrightarrow{A}.\overrightarrow{C} = 2602.55$

(b)

$\overrightarrow{A}\times \overrightarrow{C}=\left ( 140.5\widehat{i}-65.5\widehat{j} \right )\times \left ( 41.6\widehat{i}+49.5\widehat{j} \right )$