The average mean life at a radio nuclide which decays by parallel path is- AB- - 1-1-8- 10-2sec-1 2AB- - 2-1-8- 10-3sec-1

The correct option is C 50 secThe average (mean ) life is 1λ _1+λ _2=11.8× 10^ 2sec^ 1+ 1.8× 10^ 3sec^ 1=50 #160; sec . ...

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Standard XII

Physics

Alpha Decay

The average m...

Question

The average (mean) life at a radio nuclide which decays by parallel path is:
$A λ_{1} - \to B; λ_{1} = 1.8 \times 10^{- 2} s e c^{- 1}$
$2 A λ_{2} - \to B; λ_{2} = 1.8 \times 10^{- 3} s e c^{- 1}$

52.63 sec

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500 sec

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50 sec

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Solution

The correct option is C 50 sec
The average (mean ) life is $\frac{1}{λ_{1} + λ_{2}} = \frac{1}{1.8 \times 10^{- 2} s e c^{- 1} + 1.8 \times 10^{- 3} s e c^{- 1}} = 50 s e c$ .

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The average (mean) life at a radio nuclide which decays by parallel path is:Aλ1−→B;λ1=1.8×10−2sec−12Aλ2−→B;λ2=1.8×10−3sec−1

The correct option is C 50 secThe average (mean ) life is 1λ1+λ2=11.8×10−2sec−1+1.8×10−3sec−1=50sec.

The average (mean) life at a radio nuclide which decays by parallel path is:
$A λ_{1} - \to B; λ_{1} = 1.8 \times 10^{- 2} s e c^{- 1}$
$2 A λ_{2} - \to B; λ_{2} = 1.8 \times 10^{- 3} s e c^{- 1}$

The correct option is C 50 sec
The average (mean ) life is $\frac{1}{λ_{1} + λ_{2}} = \frac{1}{1.8 \times 10^{- 2} s e c^{- 1} + 1.8 \times 10^{- 3} s e c^{- 1}} = 50 s e c$ .