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. |
Calculate the amount of energy set free by the annigilation of an electron and a positron. Given that mass of electron `= 0.00055` amu and positron = 0.00055 amu. |
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Answer» Correct Answer - 1.0211 MeV |
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| 2. |
The nucleus `._(92)^(235)U` decays according to `._(92)^(238)U rarr ._(92)^(234)Th + ._(2)^(4)He`. Calculate the kinetic energy of the emitted `alpha`-particle `._(92)^(238)U = 3.85395 xx 10^(-25) kg " " ._(90)^(234)Th = 3.78737 xx 10^(-25) kg " " ._(2)^(4)He = 6.64807 xx 10^(-27)` |
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Answer» Correct Answer - `8.8 xx 10^(-13)J` |
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| 3. |
Calculate the energy released in kilowatt-hours when 100g of `._(3)^(7)Li` are converted into `._(2)^(4)He` by proton bombardment. Mass of `._(3)^(7)Li = 7.0183 am u`, mass of proton = 1.0081 amu. Write down the nuclear reaction. |
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Answer» Correct Answer - `._(3)^(7)Li + ._(1)^(1)H` (proton) `rarr 2 ._(2)^(4)He + Q` (energy released), 6.554 xx 10^(6) kWh` |
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| 4. |
Calculate the binding energy of the lithium atom `(._(3)^(7)Li)` from the following data: `{:("mass of proton",= 1.00759,am u,),("mass of neutron",= 1.00898,am u,),("mass of electron",= 0.00055,am u,),("mass of lithium atom",= 7.01818,am u,):}` |
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Answer» Correct Answer - 39.3 MeV |
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| 5. |
Find the minimum energy that a `gamma`-ray must have to give rise to an electron -positron pair. `{:("Mass of electron",= 0.00055 am u,),("Mass of positron",= 0.00055 am u,):}` |
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Answer» Correct Answer - 1.024 MeV |
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