Carbon Dating worksheet

1. Age (5,730-year half-life): 8,445.63 years

1. Turn the percent left into a fraction, F, so it can go into the decay formula

   $F$$=\frac{\mathrm{36\%}}{100}$$=0.36$

2. Find the mean-life τ, the average time one atom lasts: the 5,730-year half-life ÷ ln 2

   $\tau$$=\frac{\mathrm{5,730}\text{\hspace{0.17em}years}}{\text{ln}2}$$=\mathrm{8,266.64}\text{\hspace{0.17em}years}$

3. Work back to the time since death: multiply the mean-life by ln(1 ÷ F) to undo the decay

   $t$$=\mathrm{8,266.64}\times \text{ln}\left(\frac{1}{0.36}\right)$$=\mathrm{8,446}\text{\hspace{0.17em}years}$

4. Repeat with the older 5,568-year half-life, because labs report radiocarbon ages that way

   $\mathrm{t₁₄C}$$=\frac{\mathrm{5,568}}{\text{ln}2}\times \text{ln}\left(\frac{1}{0.36}\right)$$=\mathrm{8,207}\text{\hspace{0.17em}years}$

5. The two answers differ only because of the half-life used

   $\mathrm{8,446}-\mathrm{8,207}$$=239\text{\hspace{0.17em}years}$

2. Age (5,730-year half-life): 6,600.98 years

1. Turn the percent left into a fraction, F, so it can go into the decay formula

   $F$$=\frac{\mathrm{45\%}}{100}$$=0.45$

2. Find the mean-life τ, the average time one atom lasts: the 5,730-year half-life ÷ ln 2

   $\tau$$=\frac{\mathrm{5,730}\text{\hspace{0.17em}years}}{\text{ln}2}$$=\mathrm{8,266.64}\text{\hspace{0.17em}years}$

3. Work back to the time since death: multiply the mean-life by ln(1 ÷ F) to undo the decay

   $t$$=\mathrm{8,266.64}\times \text{ln}\left(\frac{1}{0.45}\right)$$=\mathrm{6,601}\text{\hspace{0.17em}years}$

4. Repeat with the older 5,568-year half-life, because labs report radiocarbon ages that way

   $\mathrm{t₁₄C}$$=\frac{\mathrm{5,568}}{\text{ln}2}\times \text{ln}\left(\frac{1}{0.45}\right)$$=\mathrm{6,414}\text{\hspace{0.17em}years}$

5. The two answers differ only because of the half-life used

   $\mathrm{6,601}-\mathrm{6,414}$$=187\text{\hspace{0.17em}years}$

3. Age (5,730-year half-life): 9,681.75 years

1. Turn the percent left into a fraction, F, so it can go into the decay formula

   $F$$=\frac{\mathrm{31\%}}{100}$$=0.31$

2. Find the mean-life τ, the average time one atom lasts: the 5,730-year half-life ÷ ln 2

   $\tau$$=\frac{\mathrm{5,730}\text{\hspace{0.17em}years}}{\text{ln}2}$$=\mathrm{8,266.64}\text{\hspace{0.17em}years}$

3. Work back to the time since death: multiply the mean-life by ln(1 ÷ F) to undo the decay

   $t$$=\mathrm{8,266.64}\times \text{ln}\left(\frac{1}{0.31}\right)$$=\mathrm{9,682}\text{\hspace{0.17em}years}$

4. Repeat with the older 5,568-year half-life, because labs report radiocarbon ages that way

   $\mathrm{t₁₄C}$$=\frac{\mathrm{5,568}}{\text{ln}2}\times \text{ln}\left(\frac{1}{0.31}\right)$$=\mathrm{9,408}\text{\hspace{0.17em}years}$

5. The two answers differ only because of the half-life used

   $\mathrm{9,682}-\mathrm{9,408}$$=274\text{\hspace{0.17em}years}$

4. Age (5,730-year half-life): 9,419.3 years

1. Turn the percent left into a fraction, F, so it can go into the decay formula

   $F$$=\frac{\mathrm{32\%}}{100}$$=0.32$

2. Find the mean-life τ, the average time one atom lasts: the 5,730-year half-life ÷ ln 2

   $\tau$$=\frac{\mathrm{5,730}\text{\hspace{0.17em}years}}{\text{ln}2}$$=\mathrm{8,266.64}\text{\hspace{0.17em}years}$

3. Work back to the time since death: multiply the mean-life by ln(1 ÷ F) to undo the decay

   $t$$=\mathrm{8,266.64}\times \text{ln}\left(\frac{1}{0.32}\right)$$=\mathrm{9,419}\text{\hspace{0.17em}years}$

4. Repeat with the older 5,568-year half-life, because labs report radiocarbon ages that way

   $\mathrm{t₁₄C}$$=\frac{\mathrm{5,568}}{\text{ln}2}\times \text{ln}\left(\frac{1}{0.32}\right)$$=\mathrm{9,153}\text{\hspace{0.17em}years}$

5. The two answers differ only because of the half-life used

   $\mathrm{9,419}-\mathrm{9,153}$$=266\text{\hspace{0.17em}years}$

5. Age (5,730-year half-life): 11,135.78 years

1. Turn the percent left into a fraction, F, so it can go into the decay formula

   $F$$=\frac{\mathrm{26\%}}{100}$$=0.26$

2. Find the mean-life τ, the average time one atom lasts: the 5,730-year half-life ÷ ln 2

   $\tau$$=\frac{\mathrm{5,730}\text{\hspace{0.17em}years}}{\text{ln}2}$$=\mathrm{8,266.64}\text{\hspace{0.17em}years}$

3. Work back to the time since death: multiply the mean-life by ln(1 ÷ F) to undo the decay

   $t$$=\mathrm{8,266.64}\times \text{ln}\left(\frac{1}{0.26}\right)$$=\mathrm{11,136}\text{\hspace{0.17em}years}$

4. Repeat with the older 5,568-year half-life, because labs report radiocarbon ages that way

   $\mathrm{t₁₄C}$$=\frac{\mathrm{5,568}}{\text{ln}2}\times \text{ln}\left(\frac{1}{0.26}\right)$$=\mathrm{10,821}\text{\hspace{0.17em}years}$

5. The two answers differ only because of the half-life used

   $\mathrm{11,136}-\mathrm{10,821}$$=315\text{\hspace{0.17em}years}$

6. Age (5,730-year half-life): 7,171.32 years

1. Turn the percent left into a fraction, F, so it can go into the decay formula

   $F$$=\frac{\mathrm{42\%}}{100}$$=0.42$

2. Find the mean-life τ, the average time one atom lasts: the 5,730-year half-life ÷ ln 2

   $\tau$$=\frac{\mathrm{5,730}\text{\hspace{0.17em}years}}{\text{ln}2}$$=\mathrm{8,266.64}\text{\hspace{0.17em}years}$

3. Work back to the time since death: multiply the mean-life by ln(1 ÷ F) to undo the decay

   $t$$=\mathrm{8,266.64}\times \text{ln}\left(\frac{1}{0.42}\right)$$=\mathrm{7,171}\text{\hspace{0.17em}years}$

4. Repeat with the older 5,568-year half-life, because labs report radiocarbon ages that way

   $\mathrm{t₁₄C}$$=\frac{\mathrm{5,568}}{\text{ln}2}\times \text{ln}\left(\frac{1}{0.42}\right)$$=\mathrm{6,969}\text{\hspace{0.17em}years}$

5. The two answers differ only because of the half-life used

   $\mathrm{7,171}-\mathrm{6,969}$$=203\text{\hspace{0.17em}years}$

7. Age (5,730-year half-life): 6,067.46 years

1. Turn the percent left into a fraction, F, so it can go into the decay formula

   $F$$=\frac{\mathrm{48\%}}{100}$$=0.48$

2. Find the mean-life τ, the average time one atom lasts: the 5,730-year half-life ÷ ln 2

   $\tau$$=\frac{\mathrm{5,730}\text{\hspace{0.17em}years}}{\text{ln}2}$$=\mathrm{8,266.64}\text{\hspace{0.17em}years}$

3. Work back to the time since death: multiply the mean-life by ln(1 ÷ F) to undo the decay

   $t$$=\mathrm{8,266.64}\times \text{ln}\left(\frac{1}{0.48}\right)$$=\mathrm{6,067}\text{\hspace{0.17em}years}$

4. Repeat with the older 5,568-year half-life, because labs report radiocarbon ages that way

   $\mathrm{t₁₄C}$$=\frac{\mathrm{5,568}}{\text{ln}2}\times \text{ln}\left(\frac{1}{0.48}\right)$$=\mathrm{5,896}\text{\hspace{0.17em}years}$

5. The two answers differ only because of the half-life used

   $\mathrm{6,067}-\mathrm{5,896}$$=172\text{\hspace{0.17em}years}$

8. Age (5,730-year half-life): 9,952.81 years

1. Turn the percent left into a fraction, F, so it can go into the decay formula

   $F$$=\frac{\mathrm{30\%}}{100}$$=0.3$

2. Find the mean-life τ, the average time one atom lasts: the 5,730-year half-life ÷ ln 2

   $\tau$$=\frac{\mathrm{5,730}\text{\hspace{0.17em}years}}{\text{ln}2}$$=\mathrm{8,266.64}\text{\hspace{0.17em}years}$

3. Work back to the time since death: multiply the mean-life by ln(1 ÷ F) to undo the decay

   $t$$=\mathrm{8,266.64}\times \text{ln}\left(\frac{1}{0.3}\right)$$=\mathrm{9,953}\text{\hspace{0.17em}years}$

4. Repeat with the older 5,568-year half-life, because labs report radiocarbon ages that way

   $\mathrm{t₁₄C}$$=\frac{\mathrm{5,568}}{\text{ln}2}\times \text{ln}\left(\frac{1}{0.3}\right)$$=\mathrm{9,671}\text{\hspace{0.17em}years}$

5. The two answers differ only because of the half-life used

   $\mathrm{9,953}-\mathrm{9,671}$$=281\text{\hspace{0.17em}years}$

9. Age (5,730-year half-life): 9,164.92 years

1. Turn the percent left into a fraction, F, so it can go into the decay formula

   $F$$=\frac{\mathrm{33\%}}{100}$$=0.33$

2. Find the mean-life τ, the average time one atom lasts: the 5,730-year half-life ÷ ln 2

   $\tau$$=\frac{\mathrm{5,730}\text{\hspace{0.17em}years}}{\text{ln}2}$$=\mathrm{8,266.64}\text{\hspace{0.17em}years}$

3. Work back to the time since death: multiply the mean-life by ln(1 ÷ F) to undo the decay

   $t$$=\mathrm{8,266.64}\times \text{ln}\left(\frac{1}{0.33}\right)$$=\mathrm{9,165}\text{\hspace{0.17em}years}$

4. Repeat with the older 5,568-year half-life, because labs report radiocarbon ages that way

   $\mathrm{t₁₄C}$$=\frac{\mathrm{5,568}}{\text{ln}2}\times \text{ln}\left(\frac{1}{0.33}\right)$$=\mathrm{8,906}\text{\hspace{0.17em}years}$

5. The two answers differ only because of the half-life used

   $\mathrm{9,165}-\mathrm{8,906}$$=259\text{\hspace{0.17em}years}$

10. Age (5,730-year half-life): 7,998.67 years

1. Turn the percent left into a fraction, F, so it can go into the decay formula

   $F$$=\frac{\mathrm{38\%}}{100}$$=0.38$

2. Find the mean-life τ, the average time one atom lasts: the 5,730-year half-life ÷ ln 2

   $\tau$$=\frac{\mathrm{5,730}\text{\hspace{0.17em}years}}{\text{ln}2}$$=\mathrm{8,266.64}\text{\hspace{0.17em}years}$

3. Work back to the time since death: multiply the mean-life by ln(1 ÷ F) to undo the decay

   $t$$=\mathrm{8,266.64}\times \text{ln}\left(\frac{1}{0.38}\right)$$=\mathrm{7,999}\text{\hspace{0.17em}years}$

4. Repeat with the older 5,568-year half-life, because labs report radiocarbon ages that way

   $\mathrm{t₁₄C}$$=\frac{\mathrm{5,568}}{\text{ln}2}\times \text{ln}\left(\frac{1}{0.38}\right)$$=\mathrm{7,773}\text{\hspace{0.17em}years}$

5. The two answers differ only because of the half-life used

   $\mathrm{7,999}-\mathrm{7,773}$$=226\text{\hspace{0.17em}years}$