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Nucleus Formed in the Decay of the Nucleus is an Isotope Or Isobar - Physics

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प्रश्न

Is the nucleus formed in the decay of the nucleus `""_11^22Na`, an isotope or isobar?

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उत्तर

The nucleus so formed is an isobar of `""_11^22Na` because the mass number is same, but the atomic numbers are different.

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2013-2014 (March) Delhi Set 1

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संबंधित प्रश्न

Obtain the binding energy (in MeV) of a nitrogen nucleus `(""_7^14"N")`, given `"m"(""_7^14"N")` = 14.00307 u.


Obtain the binding energy of the nuclei `""_26^56"Fe"` and `""_83^209"Bi"` in units of MeV from the following data:

`"m"(""_26^56"Fe")` = 55.934939 u

`"m"(""_83^209"Bi")`= 208.980388 u


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Mass of `"_10^20 Ne= 19.992397` u

Mass of `"_0^1H = 1.007825`  u

Mass of `"_0^1n = 1.008665` u


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(i) Verify this by calculating the proton separation energy Sp for 120Sn (Z = 50) and 121Sb = (Z = 51).

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Given 119In = 118.9058u, 120Sn = 119.902199u, 121Sb = 120.903824u, 1H = 1.0078252u.

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Explain the release of energy in nuclear fission and fusion on the basis of binding energy per nucleon curve.


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The binding energy per nucleon of 37Li and 24He nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction 3Li7 +1H12He4 + 2He4 + Q the value of energy Q released is:


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