Question 351:
easyWhich of the following statements is true for nuclear forces?
(1990)
Nuclear forces are the strongest forces in nature but they are strictly short-range forces, operating effectively only over distances of about 2 to 3 femtometers (fm). They do not follow the inverse square law.
Question 352:
easyThe average binding energy of a nucleon inside an atomic nucleus is about
(1989)
For most stable nuclei with intermediate mass numbers (30 < A < 170), the average binding energy per nucleon is approximately 8 MeV.
Question 353:
easyIf $ M(A, Z) $, $ M_p $ and $ M_n $ denote the masses of the nucleus $ ^A_Z X $, proton and neutron respectively in units of u (1 u = 931.5 $ MeV/c^2 $) and BE represents its binding energy in MeV, then:
(2008)
The binding energy is given by $ B.E. = \Delta m c^2 = [Z M_p + (A-Z) M_n - M(A, Z)] c^2 $. Rearranging this gives the nuclear mass $ M(A,Z) = Z M_p + (A-Z) M_n - B.E./c^2 $.
Question 354:
easyA nucleus $ ^A_Z X $ has mass represented by $ M(A, Z) $. If $ M_p $ and $ M_n $ denote the mass of proton and neutron respectively and B.E. the binding energy in MeV, then:
(2007)
Binding energy is the energy equivalent of the mass defect. The mass defect is $ \Delta m = [Z M_p + (A-Z) M_n - M(A, Z)] $. Thus, the binding energy is $ B.E. = \Delta m c^2 = [Z M_p + (A-Z) M_n - M(A, Z)]c^2 $.
Question 355:
easyA mixture consists of two radioactive materials $A_1$ and $A_2$ with half lives of 20 s and 10 s respectively. Initially the mixture has 40 g of $A_1$ and 160 g of $A_2$. The amount of the two in the mixture will become equal after:
(2012 Pre)
Amounts remaining are equal: $40(1/2)^{t/20} = 160(1/2)^{t/10}$. This gives $(1/2)^{t/20 - t/10} = 4$, which means $2^{t/20} = 4 = 2^2$. Thus, $t/20 = 2$, yielding $t = 40 s$.
Question 356:
easyA radioactive nucleus of mass M emits a photon of frequency $\nu$ and the nucleus recoils. The recoil energy will be
(2011 Pre)
The momentum of the emitted photon is $p = h\nu/c$. By conservation of momentum, the nucleus recoils with the same momentum. Recoil energy $E = \frac{p^2}{2M} = \frac{h^2\nu^2}{2Mc^2}$.
Question 357:
easyThe decay constant of a radio isotope is $\lambda$. If $A_1$ and $A_2$ are its activities at times $t_1$ and $t_2$ respectively, the number of nuclei which have decayed during the time $(t_1 – t_2)$:
(2010 Mains)
Activity is related to the number of nuclei by $A = \lambda N$. The number of nuclei at $t_1$ is $N_1 = A_1/\lambda$ and at $t_2$ is $N_2 = A_2/\lambda$. Number of nuclei decayed is $N_1 - N_2 = \frac{A_1 - A_2}{\lambda}$.
Question 358:
easyThe activity of a radioactive sample is measured as $N_0$ counts per minute at t = 0 and $N_0/e$ counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is:
(2010 Pre)
Using $A = A_0 e^{-\lambda t}$, we get $N_0/e = N_0 e^{-5\lambda}$, yielding $5\lambda = 1$ and $\lambda = 1/5$ per minute. The half-life is $T_{1/2} = \frac{\ln 2}{\lambda} = 5 \ln 2 = 5 \log_e 2$.
Question 359:
easyTwo radioactive materials $X_1$ and $X_2$ have decay constants $5\lambda$ and $\lambda$ respectively. Initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_1$ to that of $X_2$ will be $1/e$ after a time:
(2008)
The number of nuclei are $N_1 = N_0 e^{-5\lambda t}$ and $N_2 = N_0 e^{-\lambda t}$. Their ratio is $N_1/N_2 = e^{-4\lambda t} = e^{-1}$. Therefore, $4\lambda t = 1$, yielding $t = \frac{1}{4\lambda}$.
Question 360:
easyTwo radioactive substances A and B have decay constants $5\lambda$ and $\lambda$ respectively. At t = 0 they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be $(1/e)^2$ after a time interval:
(2007)
$N_A = N_0 e^{-5\lambda t}$ and $N_B = N_0 e^{-\lambda t}$. The ratio is $N_A/N_B = e^{-4\lambda t} = (1/e)^2 = e^{-2}$. This gives $4\lambda t = 2$, which implies $t = 1/2\lambda$.