Monday, October 5, 2020

#332 Let's suppose that N1 (t) is the number

Let's suppose that N1 (t) is the number - Math

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ChemistryExplain “#332 Let's suppose that N1 (t) is the number in Bridges math curriculum, Dr mather, Carnegie math, 10th maths, 10th grade math problems, Math
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Let's suppose that N1 (t) is the number of nuclei of the original radioactive nuclide (the mother) as a function of the time and that is its decay constant. Let's also suppose that: N1 (t) is the number of nuclei of the radioactive product (the daughter) as a function of time and that hy is its decay constant. The differential equations governing this

ChemistryExplain “#332 Let's suppose that N1 (t) is the number in Bridges math curriculum, Dr mather, Carnegie math, 10th maths, 10th grade math problems, Math

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Free Chegg Answer

We've: -

dN2(t) dt = -12N2(t) + A1Ni(t) = -12N2(t) + 11 No-dit

On solving,

dN2(t) dt = -12N2(t) + X1 Noc- dN2(t) + X2N2(t) = 11 Noc dt e-lit

On comparing the above equation with first-order differential equation,

dy + P(x)y= Q()

we've: -

y = N2(0), x = t, P = 12, Q = 41No-lit

Finding the integrating factor,

I = c/ Pdt = c/ Azdt =etzt

Multiplying our original differential equation by I,

Az ON2 () + 12e Azt N(t) = 4, Noel12-01) dt

On integrating both sides w.r.t. t,

\small \int (e^{\lambda _{2}t}\frac{\mathrm{d} N_{2}(t)}{\mathrm{d} t}+\lambda _{2}e^{\lambda _{2}t}N_{2}(t))dt=\int \lambda _{1}N_{0}e^{(\lambda _{2}-\lambda _{1})t}dt

On solving,

ed2t N2(t) = 11 Nge(13-11) dt

The above follows because using product rule: -

d(e^2 N2(t)) dt = dN2(t) dt + N2(t) dle121) = {21dN2(t) dN2 + 12N2(t)e 12t dt dt

Hence,

\small e^{\lambda _{2}t}N_2(t)=\int \lambda _{1}N_0e^{(\lambda _{2}-\lambda _{1})t}dt\Rightarrow e^{\lambda _{2}t}N_2(t)=\frac{\lambda _{1}}{\lambda _{2}-\lambda _{1}}N_0e^{(\lambda _{2}-\lambda _{1})t}+c

Thus,

\small N_2(t)=\frac{\lambda _{1}}{\lambda _{2}-\lambda _{1}}N_0e^{-\lambda _{1}t}+ce^{-\lambda _{2}t}

Putting t = 0,

\small N_2(0)=\frac{\lambda _{1}}{\lambda _{2}-\lambda _{1}}N_0+c=0\Rightarrow c=-\frac{\lambda _{1}}{\lambda _{2}-\lambda _{1}}N_0

Hence,

\small N_2(t)=\frac{\lambda _{1}}{\lambda _{2}-\lambda _{1}}N_0e^{-\lambda _{1}t}-\frac{\lambda _{1}}{\lambda _{2}-\lambda _{1}}N_0e^{-\lambda _{2}t}=\frac{\lambda _{1}}{\lambda _{2}-\lambda _{1}}N_0(e^{-\lambda _{1}t}-e^{-\lambda _{2}t})

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