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 - Physics

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Answer in Brief

The radionuclide 11C decays according to 

\[\ce{^11_6C -> ^11_5B + e+ + \text{v}}\] : T1/2 = 20.3 min

The maximum energy of the emitted positron is 0.960 MeV.

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

Calculate Q and compare it with the maximum energy of the positron emitted.

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Solution

The given nuclear reaction is:

\[\ce{^11_6C -> ^11_5B + e+ + \text{v}}\]

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

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

Atomic mass of `"m"(""_6^11"B") = 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 `""_6^11"C"` nucleus is given as:

`triangle "Q" = ["m'"(""_6"C"^11) - ["m'"(""_5^11"B")  + "m"_"e"]]"c"^2`   ...(1)

Where,

me = Mass of an electron or positron = 0.000548 u

c = 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 `""^11"C"` and 5 min the case of `""^11"B"`.

Hence, equation (1) reduces to:

`triangle"Q"  = ["m"(""_6"C"^11) - "m"(""_5^11"B") - 2"m"]"c"^2` are the atomic masses

Here `"m"(""_6"C"^11) and "m'(""_5^11"B")` are the atomic masses

∴ Δ Q = [11.011434 − 11.009305 − 2 × 0.000548] c2

= (0.001033 c2) u

But 1 u = 931.5 Mev/c2

∴ Δ Q = 0.001033 × 931.5 ≈ 0.962 MeV

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

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Chapter 13: Nuclei - Exercise [Page 463]

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NCERT Physics Class 12
Chapter 13 Nuclei
Exercise | Q 13.13 | Page 463
NCERT Physics Class 12
Chapter 13 Nuclei
Exercise | Q 13 | Page 463

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