Find the half-life of U^{238}, if one gram of it emits 1.24 * 10^4 \alpha - particles per second. (Avogadro's Number = 6.023 * 10^{23}).
Given,
\frac{dN}{dt} = 1.24 * 10^{4}
Half Life (T_{1/2}) = ?
But,
238 gm of U contains 6.023 * 10^{23} atoms
1 gm of U contains \frac{6.023 \; * \; 10^{23}}{238} = 2.51 * 10^{21}
⇒ N = 2.51 * 10^{21} atoms
We have,
\frac{dN}{dt} = 1.24 * 10^{4}
Half Life (T_{1/2}) = ?
But,
238 gm of U contains 6.023 * 10^{23} atoms
1 gm of U contains \frac{6.023 \; * \; 10^{23}}{238} = 2.51 * 10^{21}
⇒ N = 2.51 * 10^{21} atoms
We have,
\frac{dN}{dt} = \lambda \; N = \frac{ln(2)}{T_{1/2}} * N
T_{1/2} = \frac{0.693 \; * \; N}{\frac{dN}{dt}} = \frac{0.693 \; * \; 2.51 \; * \; 10^{21}}{1.24 \; * \; 10^4}
= 1.41 * 10^{17}\; Sec = 4.5 * 10^9\; years
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