Mail-Josselynerodriguez Texas A&M; International Google 016 11:55 PM O 88.7/1005

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Mail-Josselynerodriguez Texas A&M; International Google 016 11:55 PM O 88.7/1005/10/2016 10:54 PM Print Calculator of 21 Periodic Table 5 Mapd in which a radioactive source is placed inside a ridium-192 is one radioisotope used in bra patients body to treat cancer the tumor while lowering the risk of damage to healthy tissue. Iridium-192 is often used in the head or breast py allows the use of a higher than normal dose to be placed near Answer the following three questions (a, b, and c) based on the radioactive decay curve of iridium-192, shown below. Clck on the graph and move the mouse over a point to get values for that point Sample remaining (%) 100 90 70 60 50 30 20 10 2. 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 130 190 200 Time (days) a) if the initial sample is 5.50 g. what mass of the original iridium-192 remains after 55 days? Number continued below OPrevous ®Give Up & View SoltionCheck Answer Next Ext about us careers partners privacy policy terms of use con

Explanation / Answer

Hi! Welcome to Chegg

Use the following equations to calculate the number of half.lifes:

(1/2)number of half lifes = the decimal amount reaminig

total time elapsed/ length of half-lifes = number of half-lifes elapsed

I will use the most exact value that I can see on the graph

Sample remaining (%) = 30

Time (days) = 130

According to the data given:

(1/2)n=0.30 (The amount remaining after 130 days)

n*log 0.5 = log 0.30

n = log 0.3/log 0.5

n=1.737

Determining the length of the half-lifes:

130 days / x = 1.737

x = 130 days/1.737

b) x = 74.84 days

55 days/ 74.84 days = 0.73 (number of half-lifes elapsed)

(1/2)0.73 = 0.60 (Decimal fraction remaining

5.50 g * 0.60 = 3.30 g remaining

a) 3.30 g remaining

c) one-fourth of the sample to decay

You lose 25%, then 75% remains.

(1/2)n=0.75

n= log0.75/log 0.5

n= 0.42

74.84 days * 0.42 =31.06 days

c) 31.06 days (Observe in the graph that 75% of the sample ramaining has a value of 31 days approximately)

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