The percentage of Pu-238 that remains today is 70%. The time required for 25% of the nuclear fuel to remain is 176 years.
According to the question, we have to use the equation;
N =Noe^(-0.008 t)
Where;
N = amount of Pu-238 present at time t
No = amount of Pu-238 initially present
Now;
N/No = e^(-0.008 t)
The time elapsed between 1977 and today is 44 years
N/No = e^(-0.008 × 44)
N/No = 0.7
N = 0.7No
The amount of the nuclear fuel that remains = 0.7No
Percentage that remains = 0.7No × 100 = 70%
Hence 70% of the nuclear fuel remains today.
From;
0.693/t1/2 = 2.303/t log No/N
When N = 0.25No
t1/2 = 87.7 years
We need to find t
0.693/ 87.7 = 2.303/t log No/0.25No
0.693/ 87.7 = 1.3865/t
0.0079 = 1.3865/t
t = 1.3865/0.0079
t = 176 years
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