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Exam
NEET
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Physics
Chapter
Nuclear Physics

NEET · Physics · Nuclear Physics

At time t = 0 , a material is composed of two radioactive atoms A...

At time t=0t = 0, a material is composed of two radioactive atoms AA and BB, where NA(0)=2NB(0){N}_{A} ( 0 ) = 2{N}_{B} ( 0 ) The decay constant of both kind of radioactive atoms is λ\lambda. However, AA disintegrates to BB and BB disintegrates to CC. Which of the following figures represents the evolution of \dfrac{{N}_{B}\left(t\right)}{{N}_{B}\left(0\right)} with respect to time tt?At time t = 0 , a material is composed of two radioactive atoms A and B , where N _ A ( 0 ) = 2 N _ B ( 0 ) The decayAt time t = 0 , a material is composed of two radioactive atoms A and B , where N _ A ( 0 ) = 2 N _ B ( 0 ) The decayAt time t = 0 , a material is composed of two radioactive atoms A and B , where N _ A ( 0 ) = 2 N _ B ( 0 ) The decayAt time t = 0 , a material is composed of two radioactive atoms A and B , where N _ A ( 0 ) = 2 N _ B ( 0 ) The decayAt time t = 0 , a material is composed of two radioactive atoms A and B , where N _ A ( 0 ) = 2 N _ B ( 0 ) The decay

Options

  1. AA common emitter amplifier has a voltage gain of 50, an input impedance of 100 and an output impedence of 200 . the
  2. BA common emitter amplifier has a voltage gain of 50, an input impedance of 100 and an output impedence of 200 . the
  3. CA common emitter amplifier has a voltage gain of 50, an input impedance of 100 and an output impedence of 200 . the
  4. DA common emitter amplifier has a voltage gain of 50, an input impedance of 100 and an output impedence of 200 . the

Correct answer

C. A common emitter amplifier has a voltage gain of 50, an input impedance of 100 and an output impedence of 200 . the

Step-by-step solution

The rate at which nuclei of BB is forming, 

dNBdt=λNA−λNB\dfrac{d{N}_{B}}{dt} = \lambda{N}_{A} - \lambda{N}_{B}

dNBdt=2λNB0e−\lambdat−λNB\dfrac{d{N}_{B}}{dt} = 2\lambda{N}_{B0}{e}^{- \lambdat} - \lambda{N}_{B}

e−\lambdat(dNBdt+λNB)=2λNB0e−\lambdate\lambdat{e}^{- \lambdat}\left(\dfrac{d{N}_{B}}{dt} + \lambda{N}_{B}\right) = 2\lambda{N}_{B0}{e}^{- \lambdat}{e}^{\lambdat}

ddt(NBe−\lambdat)=2λNB0\dfrac{d}{dt}\left({N}_{B}{e}^{- \lambdat}\right) = 2\lambda{N}_{B0}

on integrating above the equation, 
NBe\lambdat=2\lambdatNB+NB0{N}_{B}{e}^{\lambdat} = 2\lambdat{N}_{B} + {N}_{{B}_{0}}
NBe\lambdat=2\lambdatNB0+NB0{N}_{B}{e}^{\lambdat} = 2\lambdat{N}_{B0} + {N}_{{B}_{0}}

NB=NB[1+2\lambdat]e−\lambdat{N}_{B} = {N}_{B}\left[1 + 2\lambdat\right]{e}^{- \lambdat}
ddtNB=0\dfrac{d}{dt}{N}_{B} = 0

−λ[1+2\lambdat]e−\lambdat+2λe−\lambdat=0- \lambda\left[1 + 2\lambdat\right]{e}^{- \lambdat} + 2\lambda{e}^{- \lambdat} = 0
NBmaxatt=12λ{N}_{Bmax}att = \dfrac{1}{2\lambda}

At time t = 0 , a material is composed of two radioactive atoms A and B , where N _ A ( 0 ) = 2 N _ B ( 0 ) The decay

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