NCERT solution Class 12 nuclei

NCERT Solutions

Class 12 Physics

Nuclei

NCERT Questions

Q.1
Q.2
Q.3
Q.4
Q.5
Q.6
Q.7
Q.8
Q.9
Q.10
Q.11
Q.12
Q.13
Q.14
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Q.22

Q.13

The radionuclide ¹¹C decays according to ¹¹₁₁C à ¹¹₅B+ e⁺ + ν; T_1/2 =20.3 min. The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: m (¹¹₆C) = 11.011434 u and m (¹¹₆B) = 11.009305 u,

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

The given nuclear reaction is:

¹¹₁₁C à ¹¹₅B+ e⁺ + ν

Atomic mass of, m (¹¹₆C) = 11.011434 u

Atomic mass of, m (¹¹₆B) = 11.009305 u

Half-life of ((¹¹₆C) nuclei, T (_1/2) = 20.3 min

Maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (∆Q) of the nuclear masses of the (¹¹₆C)

∆Q = [m (¹¹₆C) – [m (¹¹₆B) + m_e]] c² (1)

Where, m_e = Mass of an electron or positron = 0.000548 u

c = Speed of light

If atomic masses are used instead of nuclear masses, then we have to add 6 me in the case of C¹¹and 5m_e in the case of ¹¹B.

Hence, equation (1) reduces to:

∆Q = [m (¹¹₆C) – m (¹¹₆B) -2 m_e]] c²

Here, m (¹¹₆C) and m (¹¹₆B) are the atomic masses.

∆Q = [11.011434 − 11.009305 – (2 × 0.000548)] c²

= (0.001033 c²) u

But 1 u = 931.5

∆Q = (0.001033 × 931.5)

≈ 0.962 MeV

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

Nuclei Class 12 Physics NCERT Chapter 13 NCERT Solutions

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