Nuclei of a radioactive element $$X$$ are being produced at a constant rate $$K$$ and this element decays to a stable nucleus $$Y$$ with a decay constant $$\lambda$$ and half-life $${T}_{1/2}$$. At time $$t=0$$, there are $${N}_{0}$$ nuclei of the element $$X$$
The number $${N}_{Y}$$ of nuclei of $$Y$$ at $$t={T}_{1/2}$$ is
A
$$\quad K\cfrac { \ln { 2 } }{ \lambda } +\cfrac { 3 }{ 2 } \left( \cfrac { K-\lambda { N }_{ 0 } }{ \lambda } \right) $$
B
$$K\cfrac { \ln { 2 } }{ \lambda } -2\left( \cfrac { K-\lambda { N }_{ 0 } }{ \lambda } \right) $$
C
$$K\cfrac { \ln { 2 } }{ \lambda } +\cfrac { 1 }{ 2 } \left( \cfrac { K-\lambda { N }_{ 0 } }{ \lambda } \right) $$
D
$$K\cfrac { \ln { 2 } }{ \lambda } -\cfrac { 1 }{ 2 } \left( \cfrac { K-\lambda { N }_{ 0 } }{ \lambda } \right) $$
Correct option is C. $$K\cfrac { \ln { 2 } }{ \lambda } -\cfrac { 1 }{ 2 } \left( \cfrac { K-\lambda { N }_{ 0 } }{ \lambda } \right) $$
$${ N }_{ Y }=Kt+\left( \cfrac { K-\lambda { N }_{ 0 } }{ \lambda } \right) { e }^{ -\lambda t }-\cfrac { K-\lambda { N }_{ 0 } }{ \lambda } $$
Putting $$t=T_{1/2}=\dfrac{\ln2}{\lambda}$$
$$N_Y=K\cfrac { \ln { 2 } }{ \lambda } -\cfrac { 1 }{ 2 } \left( \cfrac { K-\lambda { N }_{ 0 } }{ \lambda } \right) $$