After 10 years, a 100 mg sample of argon-42 has decayed to 81 mg. estimate the half life of argon-42.
I'm on the last part
0.81=2^-10/h please help I don't know what to do from here it's kind of different
Ah, the wonders of radioactive decay and math! Well, fear not, my friend. Let me break it down for you with a touch of clownish charm.
So, we have the equation 0.81 = (2^-10)/h, where h represents the half-life of argon-42. To find the value of h, we need to isolate it on one side of the equation.
Step right up, ladies and gentlemen! It's time for some logarithms! Using the power of the log, we can solve this equation. Take the logarithm (base 2) of both sides, and we get:
log2(0.81) = log2((2^-10)/h)
Now, according to the wacky math rule, log2(2^-10) is equal to -10. So we can simplify the equation further:
log2(0.81) = -10/h
Now, let's put on our thinking caps. To get rid of this pesky logarithm, we can raise base 2 to the power of both sides:
2^(log2(0.81)) = 2^(-10/h)
Oh boy, this is getting exciting! Guess what? log2(0.81) is just a fancy way of saying "what power do I need to raise 2 to in order to get 0.81?" And after carrying out that calculation, we find that it's approximately 0.282.
So, we're left with:
2^0.282 ≈ 2^(-10/h)
This means that the power to which 2 needs to be raised to get approximately 0.282 is equal to the power to which 2 needs to be raised to get -10/h. Amazing, isn't it?
Now, let's simplify this further:
0.282 ≈ -10/h
We're almost there, my friend! Time to do a little algebraic dance. Multiply both sides by -h to get rid of that pesky denominator:
0.282(-h) ≈ -10
Oh, look at that! When we multiply a negative number (like -h) with a positive number (like 0.282), we get a negative result. Let's flip things around:
-0.282h ≈ -10
Now, let's solve for h by dividing both sides by -0.282:
h ≈ -10 / -0.282
Don't be fooled by all those negative signs - they cancel out, and we're left with:
h ≈ 35.460
Hooray! The estimated half-life of argon-42 is approximately 35.460 (whatever unit you're using). Isn't math just like juggling numbers and variables? Keep up the great work, my friend!
To estimate the half-life of argon-42, we can use the formula for exponential decay and solve for the half-life. In this case, the formula is:
A = A₀ * (1/2)^(t / h), where:
A is the final amount,
A₀ is the initial amount,
t is the time elapsed, and
h is the half-life.
In your case, A₀ is 100 mg, and A is 81 mg after 10 years. Let's substitute these values into the formula:
81 = 100 * (1/2)^(10 / h)
Now, let's simplify the equation:
81 / 100 = (1/2)^(10 / h)
To eliminate the exponent, we can take the logarithm (base 2) of both sides of the equation:
log₂(81 / 100) = log₂((1/2)^(10 / h))
Using the rule of logarithms, we can bring down the exponent:
log₂(81 / 100) = (10 / h) * log₂(1/2)
Now, divide both sides by log₂(1/2):
log₂(81 / 100) / log₂(1/2) = 10 / h
To solve for h (the half-life), we can rearrange the equation:
h = 10 / (log₂(81 / 100) / log₂(1/2))
Let's perform the calculations:
h = 10 / (log₂(81 / 100) / log₂(1/2))
h ≈ 5.52
Therefore, the estimated half-life of argon-42 is approximately 5.52 years.