Question 371:
easyA nucleus $^{m}_{n}X$ emits one $\alpha$ particle and two $\beta^-$ particles. The resulting nucleus is:
(2011 Pre)
Emission of an $\alpha$ particle reduces mass number by 4 and atomic number by 2 (result: $^{m-4}_{n-2}X'$). Emission of two $\beta^-$ particles increases atomic number by 2 while leaving mass number unchanged (result: $^{m-4}_{n-2+2}Z = ^{m-4}_{n}Z$).
Question 372:
easyThe number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter is an:
(2009)
For every $x$ alpha particles, the atomic number decreases by $2x$. For $2x$ beta particles, the atomic number increases by $2x$. The net change in atomic number Z is zero. Nuclei with the same atomic number are isotopes.
Question 373:
easyIn the nuclear decay given below: $^{A}_{Z}X \rightarrow ^{A}_{Z+1}Y \rightarrow ^{A-4}_{Z-1}B^* \rightarrow ^{A-4}_{Z-1}B$, the particles emitted in the sequence are:
(2009)
Step 1: Z increases by 1, A is constant -> $\beta$ emission. Step 2: Z decreases by 2, A decreases by 4 -> $\alpha$ emission. Step 3: Excited state ($B^*$) to ground state ($B$) with no change in Z or A -> $\gamma$ emission. The sequence is $\beta, \alpha, \gamma$.
Question 374:
easyThe half life of radium is about 1600 years. Out of 100 g of radium existing now, 25 g will remain undecayed after:
(2004)
The fraction remaining is $25/100 = 1/4 = (1/2)^2$. This means 2 half-lives have passed. The time elapsed is $2 \times 1600 = 3200$ years.
Question 375:
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 376:
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 377:
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 378:
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 379:
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 380:
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 $(1/2)^n = (1/2)^4 = 1/16$.