`U^238` decays into `Th^234` by the emission of an a-particle. There follows a chain of further radioactive decays, either by `alpha`-decay or by `beta-`decay. Eventually a stable nuclide is reached and after that, no further radioactive decay is possible. which of the following stable nuclides is the end product of the `U^238` radioactive decay chain ?A. `Pb^206`B. `Pb^207`C. `Pb^208`D. `Pb^209`

`U^238` decays into `Th^234` by the emission of an a-particle. There f

1.	`U^238` decays into `Th^234` by the emission of an a-particle. There follows a chain of further radioactive decays, either by `alpha`-decay or by `beta-`decay. Eventually a stable nuclide is reached and after that, no further radioactive decay is possible. which of the following stable nuclides is the end product of the `U^238` radioactive decay chain ?A. `Pb^206`B. `Pb^207`C. `Pb^208`D. `Pb^209`
Answer» Correct Answer - A (a) `(4n + 2)` series starts from `U^238` and its stable end product if `Pb^206`.

Discussion

No Comment Found

Related InterviewSolutions

The binding energy per nucleon number for deutron `H_(1)^(2)`and helium ` He_(2)^(4)` are `1.1 MeV` and `7.0 MeV` respectively . The energy released when two deuterons fase to form a belium nucleus `He_(2)^(4)` is ……..A. `2.2MeV`B. `23.6MeV`C. `28.0MeV`D. `30.2MeV`
The binding energy per nucleon number for deutron `H_(1)^(2)`and helium ` He_(2)^(4)` are `1.1 MeV` and `7.0 MeV` respectively . The energy released when two deuterons fase to form a belium nucleus `He_(2)^(4)` is ……..A. 20.8 MeVB. 16.6MeVC. 25.2MeVD. 23.6MeV
The binding energy per nucleon number for deutron `H_(1)^(2)`and helium ` He_(2)^(4)` are `1.1 MeV` and `7.0 MeV` respectively . The energy released when two deuterons fase to form a belium nucleus `He_(2)^(4)` is ……..
A neutron breaks into a proton and electorn. Calculate the eenrgy produced in this reaction in `m_(e) = 9 xx 10^(-31) kg, m_(p) = 1.6725 xx 10^(-27) kg, m_(n) = 1.6747 xx 10^(-27) kg, c = 3xx10^(8)m//sec`.
`A overset(lambda_(1)) rarr B overset (lambda_(2))rarr C` t=0 `N_(0)` 0 0 ` t N_(1) N_(2) N_(3)` In the above radioactive decay C is stable nucleus. Then:A. rate of decay of `A` will first increase and then decreasesB. number of nuclei of `B` first increase and then decreasesC. if `lambda_(2) gt lambda_(1)`, then activity of `B` will always be higher than activity of `A`D. if `lambda gt gt lambda_(2)`, then number of nucleus of `C` will always be less than number of nucleus of `B`.
`A overset(lambda_(1)) rarr B overset (lambda_(2))rarr C` t=0 `N_(0)` 0 0 ` t N_(1) N_(2) N_(3)` In the above radioactive decay C is stable nucleus. Then:A. rate of decay of A will first increase and then decreaseB. Number of nuclei of B will first increase and then decreaseC. if `lambda_(2) gt lambda_(1)` then activity of B will always be higher than activity of AD. If `lambda_(1) gt gt lambda_(2)` then number of nucleus of C will always be less than number of nucleus of B.
A sample has `4 xx 10^16` radioactive nuclei of half-life `10` days. The number of atoms decaying in `30` days is.A. `3.9 x 10^16`B. `5 xx 10^15`C. `10^16`D. `3.5 xx 10^16`
(a) Assume that a particular nucleus, in a radioactive sample, has a probability `rho` of decaying in the interval t to `t+Deltat`. Taking that the nucleus actually does not decay in the above interval, what is probability that it will decay in the interval t `+Deltat` to `t+2Deltat`? (b) Find the mean life of `.^(55)C_(0)` is its activity is known to decreases 4% per hour. `ln((25)/(24))=0.04`.
The decay constant `lambda` of a radioactive sample:A. Decreases with increase of external pressure and temperatureB. Increase with increase of external pressure and temperatureC. Decreases with increase of temperature and increase of pressureD. Is independent of temperature and pressure.
In a sample of radioactive material the probability that a particular nucleus will decay in next 2 hour is `8xx10^(-4)` find the half life of the radioactive sample. [Take `ln2=0.693,ln(0.9992)=-8xx10^(-4)`]

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment