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Question

# A sample contains two radioactive nuclei x and y with half-lives 2 hour and 1 hour respectively. The nucleus x−decays to y and y−decays into a stable nucleus z. At t=0 the activities of the components in the same were equal. Find the ratio of the number of the active nuclei of y at t=1 hours to the number at t=0.

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Solution

## We know, xλx−→yλy−→zt=0[rx]0[ry]0t=4[rx]t[ry]t [rx]0⇒ activity of x [ry]0⇒ activity of yGiven: At t=0, [rx]0=[ry]0 λx=0.6932⇒ Half-life for nuclei xλx=0.6931⇒ Half-life for nuclei y [Nx]0⇒ No. of active nuclei of x [Ny]0⇒ No. of active nuclei of y∴λx[Nx]0=λy[Ny]0or [Nx]0[Ny]0=λyλxNow maximum yield of y(Ny)t at t=4hr is : [Ny]t=[Nx]0.λxλy−λx[e−λx.t−e−λy.t]+[Ny]0e−λy.t =λy[Ny]0.λxλx[λy−λx][e−λx.t−e−λy.t]+[Ny]0e−λy.t =λy[Ny]0[λy−λx][e−λx.t−e−λy.t]+[Ny]0e−λy.t∴[Ny]t[Ny]0=[λyλy−λx.e−λx.t−λyλy−λxe−λy.t+e−λy.t] =⎡⎢ ⎢ ⎢ ⎢⎣0.6932×[0.6931−0.6932]×e(−0.693×42)+0.6932×[0.6931−0.6932]×e(−0.693×41)+e(−0.693×41)⎤⎥ ⎥ ⎥ ⎥⎦ =0.25−0.06+0.06 [Ny]t[Ny]0=0.25Hence, the ratio of the number of the active nuclei of y at t=4hours to the number at t=0 is 0.25.Hence answer is 4×0.25=1

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