Question 131:
easyHalf life of radioactive element is 12.5 hour and its quantity is 256 gm. After how much time its quantity will remain 1 gm:
(2001)
The fraction remaining is $1/256 = (1/2)^8$, which corresponds to 8 half-lives. Total time $t = 8 \times T_{1/2} = 8 \times 12.5 = 100$ Hrs.
Question 132:
easyThe relation between $\lambda$ and $T_{1/2}$ as ($T_{1/2} \rightarrow$ half life):
(2000)
By definition of radioactive decay, $N = N_0 e^{-\lambda t}$. At half-life $t = T_{1/2}$, $N = N_0/2$. So, $1/2 = e^{-\lambda T_{1/2}}$, which gives $\ln(2) = \lambda T_{1/2}$. Thus, $T_{1/2} = \frac{\ln 2}{\lambda}$.
Question 133:
easyThe life span of atomic hydrogen is:
(2000)
Atomic hydrogen is highly unstable and reactive. It readily combines with another hydrogen atom to form a stable $H_2$ molecule, making its life span a fraction of a second.
Question 134:
easyHalf life period of two elements are 40 minute and 20 minute respectively, then after 80 minute ratio of the remaining nuclei will be (Initially both have equal active nuclei):
(1998)
For the first element, 80 mins is 2 half-lives ($80/40$), leaving $N_0 / 2^2 = N_0/4$. For the second, 80 mins is 4 half-lives ($80/20$), leaving $N_0 / 2^4 = N_0/16$. The ratio is $(N_0/4) / (N_0/16) = 16/4 = 4 : 1$.
Question 135:
easyThe count rate of a Geiger Muller counter for the radiation of a radioactive material of half-life of 30 minutes decreases to $5 second^{-1}$ after 2 hours. The initial count rate was
(1995)
Time elapsed is 2 hours = 120 minutes. This is $120/30 = 4$ half-lives. The final count rate is $A = A_0 / 2^4 = A_0 / 16$. Given $A = 5 second^{-1}$, the initial count rate $A_0 = 5 \times 16 = 80 second^{-1}$.
Question 136:
easyThe half life of radium is 1600 years. The fraction of a sample of radium that would remain after 6400 years
(1991)
The number of half-lives is $n = 6400 / 1600 = 4$. The fraction of the sample remaining is $(1/2)^n = (1/2)^4 = 1/16$.
Question 137:
easyWhen a uranium isotope $^{235}_{92}U$ is bombarded with a neutron, it generates $^{89}_{36}Kr$, three neutrons and :
(2020)
The fission reaction is $^{235}_{92}U + ^{1}_{0}n \rightarrow ^{89}_{36}Kr + 3(^{1}_{0}n) + ^{A}_{Z}X$. Balancing $Z$: $92 + 0 = 36 + 0 + Z \Rightarrow Z = 56$. Balancing $A$: $235 + 1 = 89 + 3 + A \Rightarrow 236 = 92 + A \Rightarrow A = 144$. The product is $^{144}_{56}Ba$.
Question 138:
easyThe binding energy per nucleon of $^7_3Li$ and $^4_2He$ nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction $^7_3Li + ^1_1H \rightarrow ^4_2He + ^4_2He + Q$ the value of energy Q released is:
(2014)
Total binding energy of reactants = $(7 \times 5.60) + 0 = 39.20$ MeV. Total binding energy of products = $2 \times (4 \times 7.06) = 56.48$ MeV. Energy released $Q = BE_{products} - BE_{reactants} = 56.48 - 39.20 = 17.28$ MeV $\approx 17.3$ MeV.
Question 139:
easyThe binding energy of deuteron is 2.2 MeV and that of $^4_2He$ is 28 MeV. If two deuteron are fused to form one $^4_2He$ then the energy released is:
(2006)
Energy released $Q = \text{Binding Energy of Product} - \text{Total Binding Energy of Reactants}$. $Q = 28 - (2.2 + 2.2) = 28 - 4.4 = 23.6$ MeV.
Question 140:
easyIn the reaction $^2_1H + ^3_1H \rightarrow ^4_2He + ^1_0n$, if the binding energies of $^2_1H$, $^3_1H$, and $^4_2He$ are respectively a, b and c (in MeV), then the energy (in MeV) released in this reaction is:
(2005)
The energy released $Q$ is the difference between the total binding energy of the products and the total binding energy of the reactants. $Q = BE(^4_2He) - [BE(^2_1H) + BE(^3_1H)] = c - (a + b) = c - a - b$.