Calculate the energy required to disintegrate the nucleus of 73Li, if the binding energy nucleon is 5.8 Mev.
Given nucleus, 73Li
Binding energy per nucleon = 5.8 MeV
The number of nucleons in the given nucleus, (A+Z) = 7
The energy required to disintegrate the nucleus is nothing but the binding energy.
Thus the binding energy = Binding energy per nucleon × Number of nucleons
= 5.8 × 7
= 40.6 MeV
Fermi is the unit of ____.
Fermi is the unit of length, used to measure the nuclear distances.
1 fermi or femtometer(fm) = 1.0 ×10−15 m
The number of α particles and β particles emitted in the following radioactive decay are ____. 22486P →20880D where P and D represents parent nucleus and daughter nucleus.
Change in mass number = 224 − 208 = 16
Change in atomic number = 86 − 80= 6
Mass number decrease only during the alpha decay. For each alpha decay, the mass number decreases by 4.
Since the mass number decreases by 16, the number of alpha particles emitted is, 16/4 = 4
The number of α particles emitted = 4.
During the 4 alpha decays, the atomic number should decrease by 8 = 86 – 8 = 78.
But the atomic number of daughter nucleus is 80.
We know that, during the beta minus decay, atomic number increases by one and the mass number remains same.
In this case, the increase in atomic number during the beta minus decays is,
80 − 78 = 2
Thus, the number α particles emitted = 4
and the number of β particles emitted = 2
Which of the following statement(s) is (are) true? a. High average binding energy → High stability b. High average binding energy → Low stability c. Low average binding energy → Low stability d. Low average binding energy → High stability
Average binding energy is the binding energy per nucleon. If the average binding energy is high, it indicates high stability. If the average binding energy is low, it indicates low stability.
The average number of neutrons produced per fission of one uranium atom are ____.
The average number of neutrons produced by the fission of one uranium atom are 2.5 which intiate the further nuclear fission.
Which of the following source produces high amount of energy?
Coal, petroleum and wood are conventional sources of energy. Whereas uranium is nuclear source of energy. For same quantity of matter, nuclear sources will give million times larger energy than conventional sources.
Which of the following decay emits electromagnetic radiation?
In alpha decay 42He nucleus is emitted by the parent nucleus.
In beta plus decay, positron is emitted by the nucleus along with the neutrino.
In beta minus decay, electron is emitted by the nucleus along with the antineutrino.
In gamma decay, the unstable nucleus emits electromagnetic radiation to get the stability.
Which of the following statement(s) is(are) correct about the nuclear force?
Nuclear force is short-range and strongest attractive force. The nuclear force does not depend on charge of the particles.
The atom bomb works on the principle of ____.
Atom bomb works on the principle of uncontrolled nuclear fission reaction.
When a radioactive element undergoes a beta-plus decay, the atomic number of daughter nucleus ____.
In case beta-plus decay, a proton transforms into neutron.
AZP →A(Z−1)D + e++ν
Thus, the atomic number of daughter nucleus decrease by one and the mass number remains same
A radioactive element undergoes a gamma decay. Then the atomic number of the daughter nucleus ____
During the gamma decay, the energy of unstable radioactive nucleus emitted in the form of photons and the nucleus becomes stable. But there is no change in the atomic number and mass number of the nucleus.
Which of the following is a pair of isotones?
Isotones are the elements having same number of neutrons.
The number of neutrons in 146C are, N = 14−6 = 8
and the number of neutrons in 168O are, N = 16− 8 = 8
Thus, 146C and 168O is a pair of isotones
A radioactive element undergoes alpha decay for 2 times, beta minus decay for 4 times and gamma decay for 2 times. Then change in the atomic number of the daughter nucleus is ____.
Let us say the atomic number of the parent nucleus as Z.
We know that during the gamma decay, there is no change in the atomic number.
In alpha decay, the atomic number decreases by 2. Then for 2 alpha decays, the atomic number decreases by 4 = Z−4
In beta minus decay, the atomic number increases by one. Then for 4 beta minus decays, the atomic number increases by 4 = (Z−4) + 4=Z.
Thus the change in atomic number after 2 alpha decays, 4 beta minus decays and 2 gamma decays is zero.
A radioactive element, 22888Ra undergoes alpha decay for 4 times and beta minus decay for 3 times. Then the mass number of the daughter nucleus is ____.
The radioactive decay equation is,
22888Ra → 21283Bi + 4 42He + 3e−+ ν
In this disintegration the daughter nucleus is 21283Bi. Its mass number is 212.
The volume of nucleus is directly proportional to ____.
We know that the radius of nucleus is,
R =R0A 1/3
Then the volume of the nucleus is, V = (4/3)πR3
V = (4/3)πR03×A
Where R0is a constant.
Then V ∝ A
The mass of 146C nucleus is ____kg. (mp =1.00727,mn=1.00866u )
Given nucleus is, 146C
Mass of proton, mp=1.00727u
Mass of neutron, mn=1.00866u
Number of protons = Z = 6
Number of neutrons, N = A−Z
Mass of the nucleus = Mass of protons + Mass of neutrons.
= 6mp+ 8mn
= (6 × 1.00727) u + (8 × 1.00866) u
= 14.1129 u
But, we know that, 1 u = 1.660539 ×10−27 kg
Then the mass of the nucleus = 14.1129 × 1.660539 ×10−27kg
= 23.435 ×10−27kg
Which of the following fundamental particle was discovered by James Chadwick?
Neutron was discovered by James Chadwick.
The nucleus of lead, 20782Pb contains ____.
The given element is 20782Pb.
The general representation of an element is ZAX.
Where Z - Atomic number = Number of protons
A - Mass number
Number of neutrons, N = A−Z
In case of lead nucleus,
The number of protons, Z = 82
Mass number, A = 207
Then the number of neutrons, N = A −Z
= 207 −82 = 125
Thus, the lead nucleus contains 82 protons and 125 neutrons.
The pair of nuclei 4018Ar and 4020Ca is an example of ____.
Isobars are the elements having same mass number but different number of neutrons and protons.
In case of the given nuclei, the mass number is same. That is 40. But the number of protons and the number of neutrons are different.
Hence, the given pair of nuclei is an example of Isobars.
1 MeV is equal to ____ J.
We know that, 1 eV = 1.6 × 10−19 J
1 Mev = 106eV
1 MeV = 106 × 1.6 × 10−19J
= 1.6 × 10−13J
The moderator(s) used in nuclear reactor is(are)____.
In nuclear reactor, moderators are used to slow down the fast moving neutrons. The moderators commonly used in the nuclear reactor are water, heavy water (D2O) and graphite.
The ratio of the radii of the nuclei 6427Co and 12552Te is approximately ____.
Let us say, the atomic mass of Cobalt, A1=64 and the atomic mass of Tellurium, A2=125
We know that, R∝A1/3
Then, R1∝A11/3 and R2∝A21/3
Thus, the ratio of the radii of the nuclei of Cobalt and Tellurium is, R1/R2= (A1/A2)1/3
Substitute the values of A1and A2in the above equation.
The above equation can be written as, R1/R2= (43/53)1/3=4/5
Thus the ratio of the radii of the nuclei of Cobalt and Tellurium is 4:5
Nuclear reactor works on the principle of ____.
Nuclear reactor works on the principle of sustained controlled chain reaction.
Which of the following nuclei has high average binding energy and more stable?
If average binding energy of a nucleus is high, the nucleus is more stable. From the average binding energy curve, it is clear that the Fe-56 nucleus has high average binding energy which is equal to 8.75 MeV. Thus the Fe-56 nucleus is more stable.
India's first nuclear reactor is ____.
The first nuclear reactor in India is APSARA which went critical in 1956. Whereas CIRUS, Canada India Research U.S. is constructed in 1960.
When 23592U is bombarded with a slow moving neutron, it undergoes fission reaction according to the equation,
10n + 23592U → 14156Ba + 9236Kr + 3x + Q
where Q is the energy released in the fission and the x is ____
The given fission reaction is,
10n +23592U → 14156Ba + 9236Kr + 3x + Q
BY balancing the atomic number and mass number on the left hand side and the right hand side of the reaction, we will get the atomic number of x as zero and its mass number as 1. Thus the x represents a neutron.
The instrument used to measure the atomic masses is ____.
By using the mass spectrometer we can measure the accurate atomic masses.
If the binding energy of α-particle is 28.2 MeV. Calculate its mass defect.
Binding energy of the α-particle = 28.2 MeV
We know that,
Binding energy = ΔMc2
and the mass of 1u is equivalent to 931.5 MeV
Thus the mass defect, ΔM = 28.2/931.5
ΔM = 0.03027
Radioactivity was discovered by ____.
Henry Becquerel discovered the radioactivity while studying the fluorescence and phosphorescence of compounds irradiated with visible light
The unit(s) of radioactivity is (are) ____.
The SI unit of radioactivity is, becquerel (Bq), 1 Bq = one decay per second
But the older unit of radioactivity is curie (Ci), 1 Ci = 3.7 × 1010Bq
rutheford (rd) is another unit of radioactivity, 1 rd = 106Bq
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