The decay constant for 1-131 is 3.59 times 10^-3 h^-1. How much 1-131 remains af
ID: 1000923 • Letter: T
Question
The decay constant for 1-131 is 3.59 times 10^-3 h^-1. How much 1-131 remains after a week if the Initial mass was 15.0g? The decay constant for Sr-90 is 1237 min^-1. If after one year k is found to be 937 min^-1, what is the half-life of Sr-90? Calculate the binding energy of^55_25Mn.(^1_0n = 1.00867 u,^1_1H = 1.00782 u,^55_25Mn = 54.9381 u) Balance the following nuclear equations.^218_84Po rightarrow ? +^214_82Pb^212_83BI rightarrow^0_-1beta + ? ? rightarrow^4_2alpha +^207_81TI^9_4Be +^4_2He rightarrow ? +^1_0n ? +^4_2He rightarrow^12_6C +^6_3Li^12_6C + ? rightarrow^1_1H +^13_6 C A sample of C-14 has an activity of 10 disintegrations per minute and a half-life of 5730 yr. How many C-14 atoms are there in this sample? How many grams are there?Explanation / Answer
01. Radioactive molecules or atoms decay with time. Decay constant is the propotionality between the size of the population of the radioactive atoms and the rate at which the population decrease. As the fundemental equation we can write N=N(0) e^-€t
Where N is the final amount of radioactive sample
N(0)is the initial amount of the sample
€ is the radioactive constant and t is the time spent.
Here t = week. This means we have to convert this to hours. 7 days into 24 hours = 7×24 = 168hours. Now let's substitute the values to the above equation.
N=15 e^ -(168 × 3.59 ×10^-3)
= 8.205g
2. We can write an equation with decay constant and half life as below. T(1/2) = ln 2 / €
T(1/2) is the half life and € is the decay constant. In2=0.693
By substituting T(1/2) = 0.693/1237 = 5.6022min
3.calculating binding energy has 3 steps. 1.calculating the mass defect. This is the difference between the mass of the nucleus and sum of the mass of the nucleons. The given mass is 55. Mn has 24 protons and 55-24 = 31 nutrons.
Now lets calculate the mass of protons and nutrons.
Total mass = (1.00782×24) + (1.00867×31) = 55.45
Therefore the mass defect is 55.45 -55 = 0.45
Second step is converting mass defet into energy. 1 amu is 1.6607 × 10^-27 kg. So the kg in 0.45 amu is 0.45 × 1.6607 × 10^-27 = 0.7473 × 10^-27 kg.
Now we have to apply E=mc^2. C= velocity of light.
E= (0.7473 × 10^-27)(2.9979×10^8)^2
= 6.716 × 10^-11 J
Third step is converting the above binding energy into molar binding energy.to do this energy should be multiplied by avagadro constant.
Therefore molar binding energy = 6.716 × 10^-11 × 6.022 × 10^23 = 4.04 × 10^13 J/mol
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.