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Oliga [24]
3 years ago
14

If I now have 3g of Iodine-131, how much did I have 16 days ago?

Physics
1 answer:
qaws [65]3 years ago
7 0

Answer:

12 grams of Iodine-13 was present 16 days ago

Explanation:

Given:

Amount of  Iodine-131 sample present now  = 3g

To Find:

Amount of Iodine-131 sample present before 16 days = ?

Solution:

W know that the half life of  Iodine-131  is  8days

Let the amount of sample before 16 days be X

Then

The  X amount  Iodine-131 sample will decay ito ist half after 8 days

after 8 days

=> X \times \frac{1}{2}

=> \frac{X}{2}

after  next 8 days , i.e., 16 days

=> \frac{X}{2}\times \frac{1}{2}

=> \frac{X}{4}

Now we know that after 16 days we have 3 grams of Iodine-131 sample

So,

\frac{X}{4}  = 3

X = 4 \times 3

X = 12

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Answer:

36.408cm3

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The mass of an object on earth is 20 kg what will be the mass of that object on moon? And why? ​
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Suppose that you have a 680 Ω, a 720 Ω and a 1.20 kΩ resistor. (a) What is the maximum resistance you can obtain by combining th
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Explanation:

As the given data is as follows.

    R_{1} = 680 \ohm ohm\ohm,    R_{2} = 720 \ohm ohm,

   R_{3} = 1.2 k\ohm = 1200 \ohm   (as 1 k ohm = 1000 m)

(a)   We will calculate the maximum resistance by combining the given resistances as follows.

      Max. Resistance = R_{1} + R_{2} + R_{3}

                                  = (680 + 720 + 1200) \ohm ohm

                                  = 2600 ohm

or,                               = 2.6 k\ohm ohm

Therefore, the maximum resistance you can obtain by combining these is 2.6 k\ohm ohm.

(b)   Now, the minimum resistance is calculated as follows.

      Min. Resistance = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}

                                 = \frac{1}{680} + \frac{1}{720} + \frac{1}{1200}

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Hence, we can conclude that minimum resistance you can obtain by combining these is 3.683 \times 10^{-3} ohm.

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