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The Radionuclide 11c Decays According to the Maximum Energy of the Emitted Positron is 0.960 Mev.Calculate Q And Compare It with the Maximum Energy of the Positron Emitted - CBSE (Science) Class 12 - Physics

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Question

The radionuclide 11C decays according to `""_6^11C ->  _5^11B + e^+ + v:"`   `T_(1/2)` = 20.3 min

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values: `m(""_6^11C) = 11.011434 u and m(""_6^11B) = 11.009305 u`

calculate and compare it with the maximum energy of the positron emitted

Solution

The given nuclear reaction is:

`""_6^11C -> _5^11B + e^+ v`

Half life of `_6^11C` nuclei `T_(1/2)` = 20.3 min

Atomic mass of `m(""_6^11C)`= 11.011434 u

Atomic mass of `m(""_6^11B) = 11.009305 u`

Maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (ΔQ) of the nuclear masses of the `""_6^11C` nucleus is given as:

`triangleQ = [m'(""_6C^11) - [m'(""_5^11B)  + m_e]]c^2`   ...(1)

Where,

me = Mass of an electron or positron = 0.000548 u

= Speed of light

m’ = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 me in the case of `""^11C` and 5 min the case of `""^11B`.

Hence, equation (1) reduces to:

`triangleQ  = [m(""_6C^11) - m(""_5^11B) - 2m]c^2` are the atomic masses

Here `m(""_6C^11) and m(""_5^11B)` are the atomic masses

∴ΔQ = [11.011434 − 11.009305 − 2 × 0.000548] c2

= (0.001033 c2) u

But 1 u = 931.5 Mev/c2

∴ΔQ = 0.001033 × 931.5 ≈ 0.962 MeV

The value of Q is almost comparable to the maximum energy of the emitted positron.

  Is there an error in this question or solution?

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Solution The Radionuclide 11c Decays According to the Maximum Energy of the Emitted Positron is 0.960 Mev.Calculate Q And Compare It with the Maximum Energy of the Positron Emitted Concept: Radioactivity - Law of Radioactive Decay.
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