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barxatty [35]
3 years ago
12

In a photoelectric experiment, you shine light onto an electrode and record a current of 25 μA. When you apply +500 mV to the el

ectrode, the current drops to 19 μA. What is the stopping potential magnitude in V?
Physics
1 answer:
kkurt [141]3 years ago
7 0

Answer:

2.083 V.

Explanation:

Stopping potential is the potential that is required to stop the current to zero . This potential is applied externally to oppose the potential created by the photoelectric effect . It gives the measure the photoelectric potential being generated .

Here current drops to 25 μA to 19 μA by a potential of 500mV

Change in current

= 25 - 19 = 6 μA

Voltage requirement for unit reduction in current

= 500 / 6 μA

To reduce current 0f 25 μA

requirement of V = (500 / 6 )  x 25 =   2083.33 mV = 2.083 V.

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An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is W/m². What is the rms
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I think your question should be:

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is

S = 1.23*10^9 W/m^2

What is the rms value of (a) the electric field and

(b) the magnetic field in the electromagnetic wave emitted by the laser

Answer:

a) 6.81*10^5 N/c

b) 2.27*10^3 T

Explanation:

To find the RMS value of the electric field, let's use the formula:

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Therefore

E_r_m_s = sqrt*{(1.239*10^9W/m^2) / [(3.00*10^8m/s)*(8.85*10^-^1^2C^2/N.m^2)]}

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