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2 years ago

A sample contains 16 g of a radioactive isotope. how much radioactive

Physics

2 answers:

mafiozo [28]2 years ago

7 0

After one half-life, 8 g of radioactive isotope will remain in the sample.

<h3>What is radioactivity?</h3>

The act of producing radiation spontaneously is known as radioactivity. This is accomplished by an unstable atomic nucleus that want to give up some energy in order to move to a more stable form.

The following formula is used to compute the number of half lives elapsed:

$\rm N=\frac{N_0}{2^n} \\\\ N=\frac{16}{2} \\\\ N= 8 \ gram$

Hence,8 gram of radioactive isotope remains in the sample after 1 half-life.

To learn more about the radioactivity, refer to the link;

brainly.com/question/1770619

#SPJ1

Basile [38]2 years ago

5 0

Answer:

Answer is below

Explanation:

8 g

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Answer:

The lenses with different focal length are four.

Explanation:

Given that,

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We know ,

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When, R₁= 4, R₂ = 8

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$\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})$

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}+\dfrac{1}{8})$

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Put the value into the formula

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$\dfrac{1}{f}=-\dfrac{3}{40}$

$f=-13.33\ cm$

When , R₁= 4, R₂ = -8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}-\dfrac{1}{8})$

$\dfrac{1}{f}=\dfrac{3}{40}$

$f=13.33\ cm$

When , R₁= -4, R₂ = -8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}-\dfrac{1}{8})$

$\dfrac{1}{f}=-\dfrac{9}{40}$

$f=-4.44\ cm$

Hence, The lenses with different focal length are four.

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