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Softa [21]
3 years ago
5

A Si sample contains 1016 cm-3 In acceptor atoms and a certain number of shallow donors, the In acceptor level is 0.16 eV above

Ev , and EF is 0.26 eV above Ev at 300 K. How many (cm-3) In atoms are un-ionized (i.e., neutral)

Engineering
2 answers:
matrenka [14]3 years ago
6 0

Answer:

Explanation: see attachment below

creativ13 [48]3 years ago
5 0

Answer:

6.5 × 10¹⁵/ cm³

Explanation:

Thinking process:

The relation N_{o} = N_{i} * \frac{E_{f}-E_{i}  }{KT}

With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹

and ni = 1.5 × 10¹⁰

Temperature, T = 300 K

K = 1.38 × 10⁻²³

This generates N₀ = 1.654 × 10¹⁶ per cube

Now, there are 10¹⁶ per cubic centimeter

Hence, N_{d}  = 1.65*10^{16}  - 10^{16} \\           = 6.5 * 10^{15} per cm cube

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Answer:

Option (d) 2 min/veh

Explanation:

Data provided in the question:

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thus,

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or

Average time spent by the vehicle = \frac{1}{60(1-0.5)}

or

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or

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7 0
3 years ago
The temperature of a system rises by 10 °C during a heating process. Express the rise in temperature of K, R, and °F.
Lorico [155]

Explanation:

Given T = 10 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

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T = (10 + 273.15) K = 283.15 K

<u>T = 283.15 K </u>

The conversion of T( °C) to T(F) is shown below:

T (°F) = (T (°C) × 9/5) + 32  

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T (°F) = (10 × 9/5) + 32 = 50 °F

<u>T = 50 °F</u>

The conversion of T( °C) to T(R) is shown below:

T (R) = (T (°C) × 9/5) + 491.67

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T (R) = (10 × 9/5) + 491.67 = 509.67 R

<u>T = 509.67 R</u>

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Answer:

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Explanation:

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Answer:

A&C

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