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Scilla [17]
3 years ago
6

1. In a certain semiconducting material the charge carriers each have a charge of 1.6 x 10-19 C. How many are entering the semic

onductor per second when the current is 2.0 nA?
Physics
1 answer:
Gennadij [26K]3 years ago
5 0

Answer:

The number of charges is 1.25 × 10¹⁰

Explanation:

Current is the amount of charge flowing through a conductor per second. The formula for current (I) is given as:

I = Q/t

Where Q is the charge flowing in coulombs and t is the time taken in seconds.

Given that I = 2.0 nA = 2 × 10⁻⁹ A and t = 1 sec

I = Q / t

Q = It =  2 × 10⁻⁹ × 1 =  2 × 10⁻⁹ C

Since each charge = 1.6 x 10⁻¹⁹ C, therefore:

The number of charges = 2 × 10⁻⁹ C / 1.6 x 10⁻¹⁹ C = 1.25 × 10¹⁰

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PlROCA

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Two bodies, with heat capacitiesC1andC2(assumed independent oftemperature) and initial temperatureT1andT2, respectively, are pla
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3 years ago
Choose the correct statement of Kirchhoff's voltage law.
frutty [35]

Answer:

A) If one travels around a closed path adding the voltages for which one enters the negative reference and subtracting the voltages for which one enters the positive reference, the total is zero.

Explanation:

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3 years ago
In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.
MA_775_DIABLO [31]

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X \rightarrow X' + \alpha

This means that in the decay:

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- The original nucleus loses 2 nucleons (2 protons and 2 neutrons) --> so its mass number A decreases by 4 units

In this example, the original nucleus is Ac (Actinium), with

Z = 89

A = 227

After the decay, it must be

Z - 2 = 89 - 2 = 87

A - 4 = 227 - 4 = 223

We see from the periodict table, Z=87 corresponds to Francium (Fr), so the final nucleus will be francium-223 (the isotope of francium with 223 nucleons).

B) Polonium-211

In a beta-minus decay, a neutron in the nucleus turns into a proton, emitting a fast-moving electron (the beta particle) and an anti-neutrino.

n \rightarrow p + e^- + \bar{\nu}

Therefore, in this process:

- The original nucleus gains 1 protons, so its atomic number Z increases by 1 unit

- The original nucleus does not lose/gain nucleons, so its mass number A remains the same

In this example, the original nucleus is Bi (bismuth)-211, with

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So After the decay, it will be

Z + 1 = 83 + 1 = 84

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So, the nucleus will be Polonium (Z=84), isotope with 211 nucleons.

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p \rightarrow n + e^+ +\nu

Therefore, in this process:

- The original nucleus loses 1 protons, so its atomic number Z decreases by 1 unit

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In this example, the original nucleus is Na (sodium)-22, with

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So, the nucleus will be Neon (Z=10), isotope with 22 nucleons.

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X' \rightarrow X + \gamma

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Reptile [31]

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