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Mrrafil [7]
3 years ago
13

Brittle- would break when struck with force * A-Metal B-Nonmetal C-Metalloid

Physics
2 answers:
m_a_m_a [10]3 years ago
7 0
<h2>Metalloids are usually too brittle to have any structural uses.</h2>

bazaltina [42]3 years ago
4 0

-B because metal hardly breaks but non metal items such as glass or plastic does!

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A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the
Maurinko [17]

Answer:

(a). The decay constant is 1.55\times10^{-5}\ s^{-1}

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(b). The value of N₀ is 2.38\times10^{11}\ nuclei

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Explanation:

Given that,

Activity R_{0}=10\ mCi

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R=R_{0}e^{-\lambda t}

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\lambda=1.55\times10^{-5}\ s^{-1}

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Using formula of half life

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}

Put the value into the formula

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}

T_{\dfrac{1}{2}}=44.719\times10^{3}\ s

T_{\dfrac{1}{2}}=11.3\ hr

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N_{0}=\dfrac{3.70\times10^{6}}{\lambda}

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N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}

N_{0}=2.38\times10^{11}\ nuclei

(c). We need to calculate the sample's activity

Using formula of activity

R=R_{0}e^{-\lambda\times t}

Put the value intyo the formula

R=10e^{-(1.55\times10^{-5}\times30\times3600)}

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Hence, (a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

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