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3 years ago

What current, in units of milliAmperes, would produce a potential difference of 4.5

Physics

1 answer:

Bess [88]3 years ago

5 0

Answer;

<u>= 56.96 mA</u>

Explanation;

From the Ohm's law;

V = IR, where V is the potential difference in volts, I is the electric current in amperes and R is the resistance in ohms.

V = 4.5 volts, R = 79 Ω

Making I the subject of the formula we get;

I =V/R

= 4.5 /79

= 0.05696 A

But, 1 A = 1000 mA

Thus; 0.05696 × 1000

<u>= 56.96 mA</u>

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Answer:

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Explanation:

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Which of the following metals require ultraviolet light to exhibit the photoelectric effect?The options available: a. Cs, work f

Eddi Din [679]

Answer:

b. AG, work function=4.74eV

Explanation:

Ultraviolet light starts at the end of the visible light spectrum, where violet light ends:

$\lambda=380 nm =3.8\cdot 10^{-7}m$ (wavelength of lowest-energy ultraviolet light)

So, the lowest energy of ultraviolet light can be found by using the formula

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

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Substituting,

$E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.8\cdot 10^{-7} m}=5.23\cdot 10^{-19}J$

And keeping in mind that

$1 eV = 1.6\cdot 10^{-19}J$

This energy converted into electronvolts is

$E=\frac{5.23\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=3.27 eV$

The work function of a metal is the minimum energy needed to extract a photoelectron from the surface of the metal. Therefore, the metals that exhibit photoelectric effect are the ones whose work function is larger than the energy we found previously, so:

b. AG, work function=4.74eV

Because for all the other metals, visible light will be enough to extract photoelectrons.

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