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At the times zero there will be 10.0 g of tungston-187 if the half life is 23.9 hours how much will be present at the end of exactly one day

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At the end of exactly one day, there will be approximately 5.01 grams of tungsten-187 remaining, given the half-life of 23.9 hours.

The decay of a radioactive substance follows an exponential decay equation given by:


\[ N(t) = N_0 * \left((1)/(2)\right)^{\frac{t}{T_{\text{1/2}}}} \]

where:

-
\( N(t) \) is the quantity of the substance at time
\( t \),

-
\( N_0 \) is the initial quantity of the substance,

-
\( T_{\text{1/2}} \) is the half-life of the substance.

In this case, for tungsten-187,
\( N_0 = 10.0 \) g and
\( T_{\text{1/2}} = 23.9 \) hours. After one day
(\( t = 24 \) hours), we can substitute these values into the equation:


\[ N(24) = 10.0 * \left((1)/(2)\right)^{(24)/(23.9)} \]

Calculating the quantity of tungsten-187
(\(N(24)\)) at the end of exactly one day using the exponential decay equation:


\[ N(24) = 10.0 * \left((1)/(2)\right)^{(24)/(23.9)} \]


\[ N(24) = 10.0 * \left((1)/(2)\right)^(1.00209) \]


\[ N(24) \approx 10.0 * 0.500786 \]


\[ N(24) \approx 5.00786 \, \text{g} \]

So, at the end of exactly one day, there will be approximately 5.01 grams of tungsten-187 remaining, given the half-life of 23.9 hours.

User Mohamed AL ANI
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