Question 141:
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.
Question 142:
easyA radioactive elements emits one $\alpha$ and $\beta$ particle then mass number of daughter element is:
(1999)
The emission of one $\alpha$ particle reduces the mass number by 4. The emission of a $\beta$ particle does not change the mass number. Therefore, the net effect is that the mass number is decreased by 4.
Question 143:
easyFor nuclear reaction $_{92}U^{235} + _{0}n^{1} \rightarrow _{56}Ba^{144} + …… + 3 _{0}n^{1}$:
(1998)
Let the unknown product be $^{A}_{Z}X$. Balancing the atomic numbers: $92 + 0 = 56 + Z + 0 \Rightarrow Z = 36$. Balancing the mass numbers: $235 + 1 = 144 + A + 3 \Rightarrow A = 236 - 147 = 89$. The element with $Z=36$ is Krypton (Kr), so the product is $_{36}Kr^{89}$.
Question 144:
easyWhich of the following is used as a moderator in nuclear reaction?
(1997)
Moderators are used in nuclear reactors to slow down fast neutrons to thermal energies. Heavy water ($D_2O$) and graphite are commonly used as moderators. Cadmium is typically used for control rods.
Question 145:
easyThe binding energies per nucleon for a deuteron and an $\alpha$-particle are $x_1$ and $x_2$ respectively. The energy Q released in the reaction $^2_1H + ^2_1H \rightarrow ^4_2He + Q$, is
(1995)
Total binding energy of two deuterons = $2 \times (2x_1) = 4x_1$. Total binding energy of the $\alpha$-particle = $4x_2$. The energy released $Q = BE_{product} - BE_{reactants} = 4x_2 - 4x_1 = 4(x_2 - x_1)$.
Question 146:
easyIn a radioactive decay process, the negatively charged emitted $\beta$-particles are:
(2007)
In $\beta^-$ decay, a neutron inside the nucleus transforms into a proton, emitting an electron (the $\beta$-particle) and an antineutrino. Thus, they are electrons produced as a result of the decay of neutrons.
Question 147:
easyThe most penetrating radiation out of the following are
(1997)
$\gamma$-rays have the shortest wavelength and highest energy among the given options, which gives them the maximum penetrating power through matter.
Question 148:
easy75. What is the respective number of $\alpha$ and $\beta$ particles emitted in the following radioactive decay? (1995)n$^{200}_{90}X \rightarrow ^{168}_{80}Y$
Change in mass number $\Delta A = 200 - 168 = 32$. Since each $\alpha$ particle reduces $A$ by 4, number of $\alpha$ particles = $32 / 4 = 8$. Expected $Z$ after 8 $\alpha$ emissions = $90 - 8(2) = 74$. Actual final $Z$ is 80. The increase in $Z$ by 6 requires the emission of 6 $\beta$ particles.
Question 149:
easyThe nucleus $_{6}C^{12}$ absorbs an energetic neutron and emits a beta-particle ($\beta$). The resulting nucleus is (1990)
Absorption of a neutron: $_{6}C^{12} + _{0}n^{1} \rightarrow _{6}C^{13}$. Subsequent $\beta$ decay increases the atomic number by 1: $_{6}C^{13} \rightarrow _{7}N^{13} + _{-1}e^{0} + \bar{\nu}$. The resulting nucleus is $_{7}N^{13}$.
Question 150:
easyThe nucleus $^{115}_{48}Cd$, after two successive $\beta$-decay will give
(1988)
In $\beta$-decay, the atomic number $Z$ increases by 1 while the mass number $A$ remains constant. Two successive $\beta$-decays will increase $Z$ by 2. New $Z = 48 + 2 = 50$, and $A = 115$. The element with $Z=50$ is Tin (Sn), so the nucleus is $^{115}_{50}Sn$.