The half-life of potassium-40 is 1.3 billion years. If a rock specimen contained 1g of
potassium-40 when it was formed, how much potassium-40 would be left if the rock is 2.6
billion years old?

Answers

Answer 1

Answer:

2.6 billion years

Explanation:

There are essentially two ways of solving nuclear half-life problems. One way is by applying the half-life formula, which is

A

(

t

)

=

A

0

(

t

)

(

1

2

)

t

t

1

2

, where

A

(

t

)

- the quantity that remains and has not yet decayed after a time t;

A

0

(

t

)

- the initial quantity of the substance that will decay;

t

1

2

- the half-life of the decaying quantity;

In this case, the rock contains

1/4th

of the orignal amount of potassium-40, which means

A

(

t

)

will be equal to

A

0

(

t

)

4

. Plug this into the equation above and you'll get

A

0

(

t

)

4

=

A

0

(

t

)

(

1

2

)

t

t

1

2

, or

1

4

=

(

1

2

)

t

t

1

2

This means that

t

t

1

2

=

2

, since

1

4

=

(

1

2

)

2

.

Therefore,

t

=

2

t

1

2

=

2

1.3 = 2.6 billion years

A quicker way to solve this problem is by recognizing that the initial amount of the substance you have is halved with the passing of each half-life, or

t

1

2

.

This means that you'll get

A

=

A

0

2

after the first 1.3 billion years

A

=

A

0

4

after another 1.3 billion years, or

2

1.3 billion

A

=

A

0

8

after another 1.3 billion years, or

2

(

2

1.3 billion

)


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Answers

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