The isotope 125B having a mass 12-014 u undergoes - decay to 126C- 126C has an excited state of the nucleus 126C- at 4-041 MeV above its ground state- If 125B decay to 126C- the maximum kinetic energy of the - particle in units of MeV nearly is integer only-Take - m126C-12-0 u1u-931-5 MeV-c2-Where c is the speed of light in vacuum

Given, m(^12_5B)=12.014 u m(^12_6C)=12.0 u E_excitation=4.041 MeV ^12_5B → ^12_6C+β^ +γ^ +Q Q=Δ mc^2=[m(^12_5B) m(^12_6C^*)]c^2 =[12.014 12.0]c^2 =0.014× 93 ...

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Byju's Answer

Standard XII

Chemistry

Nuclear Stability

The isotope 1...

Question

The isotope $_{5}^{12} B$ having a mass $12.014 u$ undergoes $β -$ decay to $_{6}^{12} C ._{6}^{12} C$ has an excited state of the nucleus $(_{6}^{12} C^{})$ at $4.041 MeV$ above its ground state. If $_{5}^{12} B$ decay to $_{6}^{12} C^{},$ the maximum kinetic energy of the $β -$ particle in units of $MeV$ nearly is (integer only).

Take : $m (_{6}^{12} C) = 12.0 u (1 u = 931.5 MeV / c^{2}, Where c is the speed of light in vacuum)$

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Solution

Given, $m (_{5}^{12} B) = 12.014 u m (_{6}^{12} C) = 12.0 u E_{e x c i t a t i o n} = 4.041 MeV$

$_{5}^{12} B \to_{6}^{12} C + β^{-} + γ^{-} + Q$

$Q = Δ m c^{2} = [m (_{5}^{12} B) - m (_{6}^{12} C^{*})] c^{2}$

$= [12.014 - 12.0] c^{2}$

$= 0.014 \times 931.5 = 13.041 MeV$

Now, $K E_{m a x} = Q - E_{e x c i t a t i o n}$

$K E_{m a x} = 13.041 - 4.041 = 9 MeV$

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Given, m(125B)=12.014 um(126C)=12.0 uEexcitation=4.041 MeV 125B → 126C+β−+γ−+Q Q=Δmc2=[m(125B)−m(126C∗)]c2 =[12.014−12.0]c2 =0.014×931.5=13.041 MeV Now, KEmax=Q−Eexcitation KEmax=13.041−4.041=9 MeV