Question

9.) The isotope chromium-51 has a half-life of 28 days. If the original sample had a 51Cr mass of 32.0 grams, what mass of 51Cr would remain after 84 days

Answer 1

Answer: The mass of Cr-51 present at 84days = 4.00g

Explanation: Using the relation

Nt/No = e -( kt)

Where

Nt = mass remain at time t(days)=?

No= initial amount at time zero= 32.0g

K= decay constant

t = 84 days

T1/2 = Half life= 28days

k= In 2/ (T1/2)

In 2 =0.693

Which implies

k = 0.693/28

k= 0.02475 per day

Nt = No e-(Kt)

Nt = 32 e- ( 0.02475 × 84)

Nt = 32 e-(2.079)

Nt = 32 × 0.125

Nt=4.00g

Therefore the mass of Cr-51 remaining after 84days is 4.00g

Answer 2

`\large \boxed{\text{4.0 g}}`

Step-by-step explanation:

The half-life of chromium-51 (28 da) is the time it takes for half of it to decay.

After one half-life, half of the original amount will remain.

After a second half-life, half of that amount will remain, and so on.

We can construct a table as follows:

`\begin{array}{ccccc}\textbf{No. of} & &\textbf{Fraction} &\textbf{Mass}\\ \textbf{Half-lives} & \textbf{t/da} &\textbf{Remaining}&\textbf{Remaining/g}\\0 & 0 & 1 &32.0\\\\1 & 28 & (1)/(2) & 16.0\\\\2 & 56 & (1)/(4) & 8.00\\\\3 & 84 & (1)/(8) & 4.00\\\\4 & 112 & (1)/(16) & 2.00\\\\\end{array}`

We see that, after 84 da (three half-lives), ⅛ of the original mass remains.

`\text{Mass remaining} = \left((1)/(2)\right)^(3) * \text{32 g}= (1)/(8)*\text{32.0 g} = \large \boxed{\textbf{4.00 g}}`