JEE Advanced 2019Paper 1
- Exam
- JEE Advanced
- Year
- 2019
- Paper
- Paper 1
- Subject
- Physics
- Chapter
- Nuclear Physics
JEE Advanced 2019 · Paper 1 · Physics · Nuclear Physics
In a radioactive sample, _ 19 ^ 40 K nuclei either decay into...
In a radioactive sample, nuclei either decay into stable nuclei with decay constant per year or into stable nuclei with decay constant per year. Given that in this sample all the stable and nuclei are produced by the nuclei only. In time years, if the ratio of the sum of stable and nuclei to the radioactive nuclei is the value of t will be : [Given ]
Options
- A
- B
- C
- D
Correct answer
D.
Step-by-step solution
Given that, decays in two stable Nuclei. Let there be active nuclei of present at and none of present of .
Where (decay constant for ) and
(decay constant for )
These form two parallel nuclear reactions.
Equivalent decay consent,
Now, number of active nuclei of left at \left(t = t\right) = NN = {N}_{0}{e}^{- \lambdat}\dfrac{{N}_{0} - N}{N} = 99 ⇒ {N}_{0} = 100N ⇒ N = \dfrac{{N}_{0}}{100}\dfrac{{N}_{0}}{100} = {N}_{0}{e}^{- \lambdat}⇒ \ln 100 = \lambdat ⇒ t = \dfrac{2\ln 10}{\lambda} = \dfrac{2 \times 2.3}{5 \times {10}^{- 10}} = 9.2 \times {10}^{10}years
Where (decay constant for ) and
(decay constant for )
These form two parallel nuclear reactions.
Equivalent decay consent,
Now, number of active nuclei of left at \left(t = t\right) = NN = {N}_{0}{e}^{- \lambdat}\dfrac{{N}_{0} - N}{N} = 99 ⇒ {N}_{0} = 100N ⇒ N = \dfrac{{N}_{0}}{100}\dfrac{{N}_{0}}{100} = {N}_{0}{e}^{- \lambdat}⇒ \ln 100 = \lambdat ⇒ t = \dfrac{2\ln 10}{\lambda} = \dfrac{2 \times 2.3}{5 \times {10}^{- 10}} = 9.2 \times {10}^{10}years