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REY [17]
3 years ago
7

Now consider the quantum theory of conduction. At low temperatures, with scattering by lattice vibrations (phonons) not yet impo

rtant,  ~ T. The residual resistivity, 0, as T -> 0 (the text calls this “I ”) is dependent on scattering by defects (impurities, . . .). If the lattice is ‘perfectly ordered’, then  can approach infinity (or at least the size of the metal.) The residual resistivity is a measure of the purity of the sample. A relatively good sample of Cu has 0=2 10-11m.a.) If the Fermi velocity, uF, of Cu is 1.57 106 m/s, and =2 10-11m, according to eq. 10-25 what is the mean free path  in units of m? (Assume Cu is monovalent, i. e. each atom contributes 1 electron to the conduction.) (show how the units come out to give units of length in the end.)
Chemistry
1 answer:
nydimaria [60]3 years ago
4 0

Answer:

jhjhh

Explanation:

jhhjhjh

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metals can be bent and others can break,

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3 years ago
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Wich substances is most likely to form in a precipitation reaction?
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Answer:

A precipitation reaction refers to the formation of an insoluble salt when two solutions containing soluble salts are combined. The insoluble salt that falls out of solution is known as the precipitate, hence the reaction's name.

Explanation:

4 0
3 years ago
List 6 uses of acids
devlian [24]

Answer: 1. Vinegar, used in the kitchen, is a liquid containing 3-6% acetic acid. It is used in pickles and in many food preparations.

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8 0
3 years ago
The radioactive decay of a certain sample produced 846 disintegrations per minute. exactly 3.00 days later, the rate of decay wa
Margarita [4]

Answer:

\boxed{\text{1.81 da}}

Explanation:

1. Calculate the decay constant

The integrated rate law for radioactive decay is 1

\ln\dfrac{A_{0}}{A_{t}} = kt

where

A₀ and A_t are the counts at t = 0 and t

k is the radioactive decay constant

\ln \dfrac{846}{269} = k \times 3.00\\\\\ln3.145 = 3.00k\\1.146 = 3.00k\\\\k =\dfrac{1.146}{3}\\\\k = \text{0.382 /da}\\

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{0.382} = \text{1.81 da}

The half-life for decay is \boxed{\textbf{1.81 da}}.

3 0
4 years ago
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