# 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

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.

#### 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.

Is there an error in this question or solution?

#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 13 Nuclei
Exercise | Q 13.13 | Page 463
NCERT Class 12 Physics Textbook
Chapter 13 Nuclei
Exercise | Q 13 | Page 463

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