Question

A mine worker discovers an ore sample containing 500 mg of radioactive material. It is discovered that the radioactive material has a half life of 1 day. Find the amount of radioactive material remaining in the sample at the beginning of the 7th day.

Answer 1

Given: A mine worker discovers an ore sample containing 500 mg of radioactive material. It is discovered that the radioactive material has a half life of 1 day.

To find: Amount of radioactive material remaining in the sample at the beginning of the 7th day.

Solution: Any radioactive material follows first order reaction.

Let the initial mass be n0, mass at beginning of seventh day be n, half life be t1 and time elapsed be t2.

Therefore, n0= 500 mg

t1 = 1 day

Since it is asked about the amount at the beginning of the 7th day, it means that 6 days has elapsed.

Therefore, t2= 6

The formula used in case of radioactive decay is given by:

$n = n0 \: ({ \frac{1}{2} )}^{ \frac{t2}{t1} }$

$= > n = 500 \: ({ \frac{1}{2} )}^{ \frac{6}{1} }$

$= > n = 500 \: ({ \frac{1}{2} )}^{ 6 }$

$= > n = 500 \: ({ \frac{1}{64} )}$

=>n = 7.8125 mg

Therefore, the amount of radioactive material remaining in the sample at the beginning of 7th day is 7.8125 mg.

Answer 2

The amount of radioactive material remaining in the sample at the beginning of the 7th day is equal to 7.81 mg.

Given:

The initial mass of radioactive material (A₀) = 500 mg

The half-life of the radioactive material : $\bold{T_{1/2}}=1\:day$  

To Find:

The amount of the radioactive material remaining in the sample at the beginning of the 7th day.

Solution:

→ The relationship between the amount of radioactive material left (A) and the time (t) is given as:

                                         $A=A_{0} (\frac{1}{2} )^{\frac{(t)}{T_{1/2} } }$

• where 'A₀' is the initial amount of the radioactive material present.
• where 'A' is the amount of radioactive material present after time 't'.

• $where\:'\bold{T_{1/2}}'\:is \:the\:half-life\:of\:the\:radioactive\:material.$

→ Beginning of the 7th day means that 6 days are already completed.

→ Now we will use the above formula to calculate the amount of radioactive material remaining at the beginning of the 7th day:

                                       $\therefore A=A_{0} (\frac{1}{2} )^{\frac{(t)}{T_{1/2} } }\\\\\therefore A=500\times (\frac{1}{2} )^{\frac{(6)}{1 } }\\\\\therefore A=500\times(\frac{1}{2} )^{6} \\\\\therefore A= 7.81 \:mg$

Therefore the amount of radioactive material remaining in the sample at the beginning of the 7th day is equal to 7.81 mg.

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