InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A radioactive nucleus undergoes a series of deacy according to the scheme. `Aoverset(alpha)rarrA_(1)overset(beta^(-))rarrA_(2)overset(alpha)rarrA_(3)overset(gamma)rarrA_(4)` If the mass number and atomic number of `A` are `180` and `172` respectively, what are these numbers for `A_(4)`.A. `56,23`B. `180,72`C. `120,52`D. `84,38` |
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Answer» Correct Answer - B `Aoverset(beta)rarrA_(1)overset(alpha)rarrA_(2)overset(alpha)rarrA_(3)` `._(72)X^(180)overset(-1e^(0))rarr._(73)X^(180)overset(._(2)He^(4))rarr._(71)X^(176)` `._(71)X^(176)overset(._(2)He^(4))rarr._(69)X^(172)` |
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| 2. |
The compound unstabel nucleus `._(92)^(236)U` often decays in accordance with the following reaction `_(92)^(236)U rarr ._(54)^(140)Xe +`_(38)^(94)Sr ` + other particles During the reaction, the uranium nucleus 'fissions' (splits) into the two smaller nuceli have higher nuclear binding energy per nucleon (although the lighter nuclei have lower total nuclear binding energies, because they contain fewer nucleons). Inside a nucleus, the nucleons (protonsa and neutrons)attract each other with a 'strong nuclear' force. All neutrons exert approxiamtely the same strong nuclear force on each other. This force holds the nuclear are very close together at intranuclear distances. In the nuclear reaction presented above, the 'otter particles' might be .A. An alpha particle, which consists of two protons and two neutronsB. two protonsC. one proton and one neutronD. two neutrons |
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Answer» Correct Answer - d Nuclear reactions conserve total charge, and also conserve the total approxiamte mass. The other particles in the reaction will have mass `=236 -140-94=2`. The other particles are two neutrons. Hence, (a) is not correct. For nuclei, number of protons tells the charege. So, the other particles must have charge `Z` such that `92 =54+38+Z` `:. Z =0` Therefore, the other particles have a total atomics mass `2` and total charge `0`. Hence, only (d) is correct. |
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| 3. |
`.^(90)Sr` decays to `.^(90)Y` by `beta` decay with a half-life of `28` years. `.^(90)Y` decays by `beta`-to `.^(90)Zr` with a half-life of `64h`. A pure sample of `.^(90)Sr` is allowed to decay. What is the valued of `(N_(Sr))/(N_(y))` after (a) `1h` (b) `10` years ? |
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Answer» `N_(sr)=N_(Sr)^(0)r^((-lambda_(sr)t))` ....(1) `N_(y)=(lambda_(sr)N_(sr)^(0))/(lambda_(y)-lambda_(sr))[e^((lambda_(sr)t))-e^((-lambda_yt))]` .......(2) Dividing the two equations `(N_(y))/(N_(sr))(lambda_(sr))/(lambda_(y)-lambda_(sr))[1-e^((-lambdat))(lambda_(sr)-lambda_(y))t]` `lambda_(sr)=0.693//(28xx365xx24)=2.825xx10^(-6)h^(-1)` `lambda_(y)=0.693//64=0.0108 h^(-1)` (a) For `t=1h` and using the values for the decay constant `N_(sr)//N_(y)=3.56xx10^(5)` (b) For `t=10 "year", N_(sr)//N_(y)=3823` |
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