Find the number of disintegrations per minute emitted by 1.1 mol of 232-Th in 1 min. The half-life of 232-Th is 1.4 x 1010 year
I was able to do this with the following technique:
1.1 x 6.022 x10^23 =6.62x10^23
1.4x10^10=7.358x10^15 minutes
I think then it is 6.62x10^23 x 7.358x10^15
But I want to learn how to do this problem using the formulas below instead of the .693/half life.
There are the following formulas:
starting amount) x (1/2)number of half-lives = ending amount (sometimes remaining rather than ending is used)
(1/2)number of half-lives is the decimal fraction which remains (1.000 is the original starting amount, 0.500 at the end of one half-life, 0.250 at the end of two, 0.125 at the end of three, etc.)
number of half-lives that have occurred can be expressed as (total time elasped ÷ length of half-life)
I think what you want applies only to what remains or what decays and not to activity.
If we want to know what remains after 2 half lives it is (1/2^n) = 1/4 or after 3 half lives is (1/2^3) = 1/8 etc.
What you have here is activity at t=0.
Ro = kNo
k = 0.693/t1/2 = 0.693/7.36E15 = about 9.4E-17
Ro = 9.4E-17 x 6.62E23 = ?dpm
To find the number of disintegrations per minute emitted by 1.1 mol of 232-Th in 1 min using the formulas you provided, follow these steps:
1. Convert the half-life of 232-Th from years to minutes. Since 1 year is equivalent to 525,600 minutes, the half-life of 1.4 x 10^10 years is equal to (1.4 x 10^10) x (525,600) minutes.
2. Divide the total time elapsed by the length of the half-life to find the number of half-lives that have occurred. In this case, since the total time elapsed is 1 minute and the half-life is (1.4 x 10^10) x (525,600) minutes, divide 1 minute by (1.4 x 10^10) x (525,600) minutes.
3. Use the formula (1/2)^(number of half-lives) to calculate the decimal fraction that remains. Raise 0.5 to the power of the number of half-lives calculated in step 2.
4. Multiply the starting amount by the decimal fraction that remains to find the ending amount. Since the initial amount is 1.1 mol, multiply 1.1 mol by the decimal fraction calculated in step 3.
The result will give you the number of disintegrations per minute emitted by 1.1 mol of 232-Th in 1 min using the given formulas.
To calculate the number of disintegrations per minute emitted by 1.1 mol of 232-Th in 1 min using the given formulas, we need to determine the number of half-lives that have occurred within 1 minute.
The number of half-lives that have occurred can be expressed as (total time elapsed ÷ length of half-life). In this case, the total time elapsed is 1 minute and the half-life of 232-Th is 1.4 x 10^10 years.
To convert years to minutes, we need to multiply by the conversion factor: 1 year = 525,600 minutes.
So, the length of half-life in minutes is (1.4 x 10^10 years) * (525,600 minutes/year) = 7.358 x 10^15 minutes.
Now we can calculate the number of half-lives that have occurred within 1 minute:
Number of half-lives = (total time elapsed ÷ length of half-life) = 1 minute ÷ (7.358 x 10^15 minutes) = 1.36 x 10^-16 half-lives.
Since each half-life results in a reduction by a factor of 0.5, the decimal fraction which remains after a given number of half-lives is (0.5)^(number of half-lives).
In this case, the starting amount is 1.1 mol, and the ending amount (remaining amount) after the number of half-lives can be calculated as follows:
Ending amount = (starting amount) x (0.5)^(number of half-lives) = 1.1 mol x (0.5)^(1.36 x 10^-16) = 1.1 mol x 0.9999999999999999 (approximately) = 1.0999999999999999 mol.
Finally, the number of disintegrations per minute emitted by 1.1 mol of 232-Th in 1 min can be calculated by subtracting the ending amount from the starting amount and multiplying by Avogadro's number (6.022 x 10^23 disintegrations/mol):
Number of disintegrations per minute = (starting amount - ending amount) x Avogadro's number = (1.1 mol - 1.0999999999999999 mol) x (6.022 x 10^23 disintegrations/mol) = 0.0000000000000001 mol x (6.022 x 10^23 disintegrations/mol) = 6.022 x 10^13 disintegrations per minute.
Therefore, the number of disintegrations per minute emitted by 1.1 mol of 232-Th in 1 min is approximately 6.022 x 10^13.