JEE Advanced 2016 Paper 1 · Physics · Atoms, Nuclei and Dual Nature of Radiation · medium

A radioactive nuclide X of half-life 8 days decays into a nuclide Y, which is itself radioactive and has a half-life of 4 days and decays to a stable…

Numerical answer

A radioactive nuclide

X

of half-life

8

days decays into a nuclide

Y

, which is itself radioactive and has a half-life of

4

days and decays to a stable nuclide. At

t=0

a sample contains only nuclei of

X

. The time, in days, at which the number of nuclei of

Y

in the sample is maximum is

Show answer and solution

Answer: 8

Let

\lambda_X=\dfrac{\ln2}{8}

and

\lambda_Y=\dfrac{\ln2}{4}

per day. The nuclei of

Y

obey

\dfrac{dN_Y}{dt}=\lambda_XN_X-\lambda_YN_Y

with

N_X=N_0e^{-\lambda_Xt}

, whose solution is

N_Y=\dfrac{\lambda_XN_0}{\lambda_Y-\lambda_X}\left(e^{-\lambda_Xt}-e^{-\lambda_Yt}\right)

. The number of

Y

nuclei is maximum when

dN_Y/dt=0

, i.e. when the activities are equal,

\lambda_XN_X=\lambda_YN_Y

, which gives

t_{max}=\dfrac{\ln(\lambda_Y/\lambda_X)}{\lambda_Y-\lambda_X}=\dfrac{\ln2}{\ln2\left(\frac14-\frac18\right)}=8

days. A numerical scan of

N_Y(t)

confirms the maximum at

t=8

days. Notice that this is not equal to a simple sum or average of the half-lives. The result also agrees with the general expression, which for

T_{1/2}(Y)=\tfrac12T_{1/2}(X)

reduces to

t_{max}=T_{1/2}(X)

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