A piece of wood from the ruins of an ancient building was found to have a 14 C activity of 12 disintegrations per minute per gram of its carbon content. The 14 C activity of the living wood is 16 disintegrations per minute per gram. How long ago did the tree, from which the wooden sample came, die? Given half-life of 14 C is 5760 yr.

This is a Short Answer Type Questions as classified in NCERT ExemplarExplanation- R= 12dis/min per gm and R0= 16dis/min per gm T1/2= 5760yrR= e - λ t e λ t By taking log λ t l o g e = l o g R 0 R λ t = log1016/12 × 2.303 t = 2.303 ( l o g 4 - l o g 3 ) λ = 2.303 (0.6020-4.771)5760/0.6931 = 2391.20yr

Ask & Answer Question

Nuclei Physics Ncert Solutions Class 12th Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

A piece of wood from the ruins of an ancient building was found to have a ¹⁴C activity of 12 disintegrations per minute per gram of its carbon content. The ¹⁴C activity of the living wood is 16 disintegrations per minute per gram. How long ago did the tree, from which the wooden sample came, die? Given half-life of ¹⁴C is 5760 yr.

2 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

4 months ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- R= 12dis/min per gm and R₀= 16dis/min per gm T_1/2= 5760yr

R= $e^{- λ t}$ $e^{λ t}$

By taking log

$λ t l o g e = l o g \frac{R_{0}}{R}$

$λ t$ = log₁₀16/12 $\times 2.303$

$t = \frac{2.303 (l o g 4 - l o g 3)}{λ}$

= 2.303 (0.6020-4.771)5760/0.6931

= 2391.20yr

0 Upvotes 0 Downvotes

Similar Questions for you

Find the mass number of an atom whose radius is half of that of a given atom of mass number 192.

alok kumar singh

$r = R_{0} {(192)}^{\frac{1}{3}}$

$\frac{r}{2} = R_{0} {(m)}^{\frac{1}{3}}$

$m = \frac{192}{8} = 24$

A nuclear fusion reaction is given by If atomic mass of He is 4.0026 amu and that of O is 15.9994 amu then the Q-value of the reaction will be approximately

alok kumar singh

Q = [4 *4.0026 – 15.9994] *931.5 MeV

Q = 10.2 MeV

A radioactive simple is undergoing a decay. At any time t₁, its activity is A and another time t₂, the activity is $\frac{A}{5} .$ What is the average life time for the sample?

alok kumar singh

$A = A_{0} e^{- λ t_{1}}$ [Radio active decay law]

$\frac{A}{5} = A_{0} e^{- λ (t_{2} - t_{1})}$

$\Rightarrow A v e r a g e l i f e = \frac{1}{λ} = \frac{(t_{2} - t_{1})}{l n 5}$

A sample contains 10^-2kg each of two substances A and B with half lives 4s and 8s respectively. The ration of their atomic weights is 1 : 2. The ratio of the amounts of A and B after16s is $\frac{x}{100}$ . The value of x is ______.

alok kumar singh

$f o r A, t_{\frac{1}{2}} = 4 s e c$

$= m A_{0} e^{-} \frac{0.693}{4} \times 16$

$m_{A} = m_{A 0} e^{- 4 \times 0.693}$ -(1)

for B,

for B, $t_{\frac{1}{2}} = 8 s e c$

$m_{B} = m_{B_{0}} e^{- 2 \times 0.693}$ -(2)

$\frac{m_{A}}{m_{B}} = \frac{m_{A O}}{m_{B O}} \frac{e^{- 4} \times 0.693}{e^{- 2} \times 0.693}$

$\frac{m_{A}}{m_{B}} = \frac{25}{100} = \frac{x}{100} x = 25$

A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6MeV while that of fragments is 8.5MeV. The total gain in the Binding Energy in the process is:

alok kumar singh

The reaction is X²? → Y¹²? + Z¹²?
Binding energies per nucleon are: X=7.6 MeV, Y=8.5 MeV, Z=8.5 MeV.
Gain in binding energy (Q) = (Binding energy of products) - (Binding energy of reactants)
Q = (120 × 8.5 + 120 × 8.5) - (240 × 7.6) MeV
Q = (2 × 120 × 8.5) - (240 × 7.6) MeV = 2040 - 1824 = 216 MeV.

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question