Question

The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be the mass (in mg) remaining after t years. Then we know the following. y ( t ) = y ( 0 ) e k t = ⋅ e k t Since the half-life is 30 years, then y ( 30 ) = mg. Thus, e 30 k = . Therefore, k = . Now, remembering that ln x n = n ln x and that e ln z = z , we have y ( t ) = 180 e ( t / 30 ) ( − ln 2 ) = mg. How much of the sample remains after 70 years? After 70 years we have the following. y ( 70 ) = 180 ⋅ 2 = mg (Round your answer to two decimal places.) After how long will only 1 mg remain? To find the time at which only 1 mg remains, we must solve 1 = y ( t ) = 180 ( 2 − t / 30 ) , and so we get the following. t = − 30 log 2 ( ) Hence, we conclude the following. t = yr (Round your final answer to one decimal place.)

Answer 1

Explanation:

1) Initial mass of the Cesium-137=$N_o$= 180 mg

Mass of Cesium after time t = N

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

Half life of the cesium-137 = $t_{1/2}=30 years[p/tex]

where,

[tex]N_o$ = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the isotope

$\lambda$ = rate constant

$N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}$

Now put all the given values in this formula, we get

$N=180mg\times e^{-\frac{0.693}{30 years}\times t}$

Mass that remains after t years.

$N=180 mg\times e^{0.0231 year^{-1}\times t}$

Therefore, the parent isotope remain after one half life will be, 100 grams.

2)

t = 70 years

$N_o=180 mg$

$t_{1/2}= 30 yeras$

$N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}$

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

3)

$N_o=180 mg$

$t_{1/2}= 30 yeras$

N = 1 mg

t = ?

$1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}$

$\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t$

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.