Question 191:
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.
Practice NEET Modern Physics Questions
Question 192:
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 193:
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 194:
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 195:
easyAn element A decays into element C by a two step process $A \rightarrow B + _{2}He^{4}; B \rightarrow C + 2e^{-}$. Then
(1989)
In alpha decay ($A \rightarrow B$), the atomic number Z decreases by 2. In two beta decays ($B \rightarrow C$), the atomic number increases by $2 \times 1 = 2$. Thus, the final atomic number of C equals the initial atomic number of A. Elements with the same Z are isotopes.
Question 196:
easyCurie is a unit of
(1989)
Curie (Ci) is a non-SI unit of radioactivity, originally defined as the activity of 1 gram of radium-226 ($3.7 \times 10^{10}$ decays per second).
Question 197:
easyA radioactive element has half life period 800 years. After 6400 years what amount will remain?
(1989)
The number of half-lives $n = \frac{6400}{800} = 8$. The fraction remaining is $(1/2)^n = (1/2)^8 = 1/256$.
Question 198:
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 199:
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. $Q = BE(^4_2He) - [BE(^2_1H) + BE(^3_1H)] = c - (a + b) = c - a - b$.
Question 200:
easyFor the given reaction, the particle X is:n$^{11}_{6}C \rightarrow ^{11}_{5}B + \beta^+ + X$
(2000)
In $\beta^+$ (positron) decay, a proton is converted into a neutron, releasing a positron ($e^+$ or $\beta^+$) and an electron neutrino ($\nu$) to conserve lepton number. Hence, particle X is a neutrino.