a)indium-115 has a half life of 4.5 hours. if you start with a 12.0-mg sample of indium-115, how much will remain after 13.5 hours?
b)how much of a 3.50-mg smaple of nickel-63 will remain after 368 years? the half-life of nickel-63 is 92 years?
There is a short way and a long way of doing these. The short way is easy if the numbers are right. In this case they are.
a) Note that 13.5/4.5 = 3.00 half lives; therefore, 1/2 of 12.0 mg sample = 6.0 mg left after the first half life, 1/2 of that = 3.0 mg left after the 2nd half life and 1/2 of that = 1.5 mg left after the 3rd half life.
b) This is done the same way if you notice that 368 years/92 years = 4.00 half lives.
Here is the long way of doing them when the number of half lives is not an even number.
k = 0.693/t1/2
ln(No/N) = kt
So, for the first one,
k = 0.693/t1/2
k = 0.693/4.5 = 0.154
Then substitute into the other equation for k you've just calculated.
ln((No/N) = kt
ln(12.0/N) = 0.154*13.5
ln(12.0/N) = 2.079
Take the antilog of both sides.
12.0/N = 8.0
N = 12.0/8.0 = 1.5 mg
The b part is done the same way the long way, too.
N=12.0/1.053 =
a) Well, if indium-115 has a half life of 4.5 hours, then after 13.5 hours, we can expect it to go through a few transformations. First, it will halve at 4.5 hours, then halve again at 9 hours, and finally, halve once more at 13.5 hours. So, after the full 13.5 hours, the remaining amount of indium-115 can be calculated using the formula N = N₀ * (1/2)^(t/t₁/₂), where N is the remaining amount, N₀ is the initial amount, t is the duration, and t₁/₂ is the half-life. Putting in the values, we get N = 12.0 mg * (1/2)^(13.5/4.5). So after these calculations, the punchline is... wait for it... approximately 1.50 mg of indium-115 will remain. That's one resilient little indium, isn't it?
b) Ah, nickel-63, it loves to take its time. With a half-life of 92 years, we can expect some serious transformations over the course of 368 years. Let's do the math. Dividing 368 by 92, we find that there have been 4 half-life periods during this time. To calculate the remaining amount, we use the formula N = N₀ * (1/2)^(t/t₁/₂). Plugging in the numbers, we find N = 3.50 mg * (1/2)^4. Now, let me carry the calculation for you...drumroll, please...and the answer is approximately 0.22 mg of nickel-63. Looks like it has really decreased, just like the chances of me ever finding a nickel on the street!
a) To determine the amount of indium-115 remaining after 13.5 hours, we'll use the formula:
N(t) = N₀ * (1/2)^(t / t₁/₂)
Where:
N(t) is the amount remaining after time t
N₀ is the initial amount
t₁/₂ is the half-life of the substance
t is the elapsed time
Given:
Initial amount (N₀) = 12.0 mg
Half-life (t₁/₂) = 4.5 hours
Elapsed time (t) = 13.5 hours
Plugging in the values into the formula:
N(t) = 12.0 mg * (1/2)^(13.5 / 4.5)
Calculating:
N(t) = 12.0 mg * (1/2)^3
N(t) = 12.0 mg * (1/8)
N(t) = 1.5 mg
Therefore, after 13.5 hours, approximately 1.5 mg of indium-115 will remain.
b) To determine the amount of nickel-63 remaining after 368 years, we'll use the same formula:
N(t) = N₀ * (1/2)^(t / t₁/₂)
Given:
Initial amount (N₀) = 3.50 mg
Half-life (t₁/₂) = 92 years
Elapsed time (t) = 368 years
Plugging in the values into the formula:
N(t) = 3.50 mg * (1/2)^(368 / 92)
Calculating:
N(t) = 3.50 mg * (1/2)^4
N(t) = 3.50 mg * (1/16)
N(t) = 0.21875 mg
Therefore, after 368 years, approximately 0.21875 mg of nickel-63 will remain.
a) To find out how much indium-115 will remain after 13.5 hours, we can use the formula for exponential decay:
Amount remaining = Initial amount * (1/2)^(time/half-life)
Given:
Initial amount of indium-115 = 12.0 mg
Half-life of indium-115 = 4.5 hours
Time = 13.5 hours
Plugging these values into the formula:
Amount remaining = 12.0 mg * (1/2)^(13.5/4.5)
Now, let's solve it step by step:
1. Calculate the exponent: 13.5 / 4.5 = 3.
2. Find the value of (1/2)^3 using a calculator: (1/2)^3 = 1/8 = 0.125.
3. Multiply the initial amount by the calculated value: 12.0 mg * 0.125 = 1.5 mg.
Therefore, after 13.5 hours, 1.5 mg of indium-115 will remain.
b) Similar to the previous question, we can use the same formula for exponential decay to find the amount of nickel-63 remaining after 368 years:
Amount remaining = Initial amount * (1/2)^(time/half-life)
Given:
Initial amount of nickel-63 = 3.50 mg
Half-life of nickel-63 = 92 years
Time = 368 years
Plugging these values into the formula:
Amount remaining = 3.50 mg * (1/2)^(368/92)
Now, let's solve it step by step:
1. Calculate the exponent: 368 / 92 = 4.
2. Find the value of (1/2)^4 using a calculator: (1/2)^4 = 1/16 = 0.0625.
3. Multiply the initial amount by the calculated value: 3.50 mg * 0.0625 = 0.21875 mg.
Therefore, after 368 years, approximately 0.21875 mg of nickel-63 will remain.