Question 291:
50. In a radioactive material the activity at time $t_1$ is $R_1$ and at a later time $t_2$, it is $R_2$. If the decay constant of the material is $\lambda$, then: (2006)
Activity follows the decay law $R = R_0 e^{-\lambda t}$. At $t_1$, $R_1 = R_0 e^{-\lambda t_1}$. At $t_2$, $R_2 = R_0 e^{-\lambda t_2}$. Dividing the two gives $R_1 / R_2 = e^{-\lambda (t_1 - t_2)}$, which rearranges to $R_1 = R_2 e^{-\lambda(t_1-t_2)}$.
Question 292:
52. A sample of radioactive element has a mass of 10 gm at an instant $t = 0$. The approximate mass of this element in the sample after two mean lives is: (2003)
Mean life is $\tau = 1/\lambda$. Mass remaining after time $t$ is $m = m_0 e^{-\lambda t}$. After two mean lives, $t = 2\tau = 2/\lambda$. So, $m = 10 e^{-2} = 10 / (2.718)^2 \approx 1.35$ gm.
Question 293:
53. A sample of radioactive element containing $4 \times 10^{16}$ active nuclei. Half life of element is 10 days, then number of decayed nuclei after 30 days: (2002)
30 days represents 3 half-lives ($30/10 = 3$). The number of active nuclei remaining is $N = N_0 (1/2)^3 = (4 \times 10^{16})/8 = 0.5 \times 10^{16}$. The number of decayed nuclei is $4 \times 10^{16} - 0.5 \times 10^{16} = 3.5 \times 10^{16}$.
Question 294:
57. The half life of a radio nuclide is 77 days then its decay constant is: (1999)
The decay constant is $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{77}$. Calculating this yields $\lambda = 0.009$/day.
Question 295:
71. A nuclear reaction given by $_{Z}X^{A} \rightarrow _{Z+1}Y^{A} + _{-1}e^{0} + \bar{\nu}$ represents: (2003)
The emission of an electron ($_{-1}e^{0}$) along with an antineutrino ($\bar{\nu}$), resulting in an increase of atomic number by 1 while the mass number remains unchanged, is the characteristic of $\beta$-decay.
Question 296:
72. A deutron is bombarded on $_{8}O^{16}$ nucleus then $\alpha$-particle is emitted then product nucleus is: (2002)
The nuclear reaction is $_{8}O^{16} + _{1}H^{2} \rightarrow _{2}He^{4} + _{Z}X^{A}$. Balancing the atomic numbers: $8 + 1 = 2 + Z \Rightarrow Z = 7$. Balancing the mass numbers: $16 + 2 = 4 + A \Rightarrow A = 14$. The product is $_{7}N^{14}$.