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Annette [7]
3 years ago
10

Consider an electrophysiology experiment in which a 0.200 ~mm0.200 mm diameter round silver wire is used as a stimulation electr

ode. In this experiment, a current of 4.69~ mA4.69 mA is applied through the wire. The density of free electrons in silver is 5.86 \times 10^{28}~ \frac{\text{electrons}}{m^3}5.86×10 ​28 ​​ ​m ​3 ​​ ​ ​electrons ​​ . Calculate the magnitude of the drift velocity of the electrons in the silver wire under these conditions. Enter your answer in units of \frac{\text{meters}}{\text{second}} ​second ​ ​meters ​​ using calculator scientific notation (e.g., 1.23e-41.23e−4 for 0.0001230.000123).

Physics
1 answer:
Kaylis [27]3 years ago
4 0

Explanation:

Below is an attachment containing the solution.

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This question relates to the practicality of searching for intelligent life in other solar systems by detecting their radio broa
Montano1993 [528]

Answer:

2.77287\times 10^{15}\ m

Explanation:

P = Power = 50 kW

n = Number of photons per second

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

\nu = Frequency = 781 kHz

r = Distance at which the photon intensity is i = 1 photon/m²

Power is given by

P=nh\nu\\\Rightarrow n=\dfrac{P}{h\nu}\\\Rightarrow n=\dfrac{50000}{6.626\times 10^{-34}\times 781000}\\\Rightarrow n=9.66201\times 10^{31}\ photons/s

Photon intensity is given by

i=\dfrac{n}{4\pi r^2}\\\Rightarrow 1=\dfrac{9.66201\times 10^{31}}{4\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{9.66201\times 10^{31}}{4\pi}}\\\Rightarrow r=2.77287\times 10^{15}\ m

The distance is 2.77287\times 10^{15}\ m

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3 years ago
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