Nuclei Test

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Nuclei Test - 10

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Weekly Quiz Competition

Question 1
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If $$M_{0}$$ is the mass of an oxygen isotope $$_{8}O^{17}$$ , Mp and $$M_{N}$$ are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is

A
$$8{M_p}c^2$$

B
$$(-M_{0}+8M_{p}+9M_{N})c^{2}$$

C
$$M_{0}c^{2}$$

D
$$(M_{0}-17M_{N})c^{2}$$

Solution

$$ BE = (8M_{p}+9M_{N} -M_{0})C^{2}$$
Question 2
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A radioactive nucleus (initial mass number A and atomic number Z) emits $$3$$ $$\alpha$$-particles and $$2$$ positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

A
$$\displaystyle \frac{\mathrm{A}-\mathrm{Z}-8}{\mathrm{Z}-4}$$

B
$$\displaystyle \frac{\mathrm{A}-\mathrm{Z}-4}{\mathrm{Z}-8}$$

C
$$\displaystyle \frac{\mathrm{A}-\mathrm{Z}-12}{\mathrm{Z}-4}$$

D
$$\displaystyle \frac{\mathrm{A}-\mathrm{Z}-4}{\mathrm{Z}-2}$$

Solution

When a radioactive nucleus emits 1 α-particle mass number decreases by 4 and the atomic number decreases by 2.

So after the emission of $$3 - \alpha$$ particles
New atomic mass, $$A' = A - 4 \times 3 = A - 12$$
New atomic number $$Z'= Z - 6$$

When this nucleus emits $$1 β$$-particle (positron), the atomic mass remains unchanged but the atomic number decreases by 1.

So after emission of 2 positrons
$$A" = A' = A-12$$
$$Z" = Z'-2×1 = Z-8$$

So, the ratio of neutrons to protons-

$$\dfrac{n}{p} = \dfrac{A-Z-4}{Z-8}$$
Question 3
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Assume that a neutron breaks into a proton and an electron. The energy released during this process is
(Mass of neutron $$=1.6725\times 10^{-27} kg$$, Mass of proton $$=1.6725\times 10^{-27} kg$$ ,Mass of electron $$=9\times 10^{-31} kg$$)

A
$$0.50625$$$$\mathrm{MeV}$$

B
$$7.10$$ $$\mathrm{MeV}$$

C
$$6.30$$ $$\mathrm{MeV}$$

D
$$5.40$$ $$\mathrm{MeV}$$

Solution

$$n\rightarrow p+ e$$
$$\Delta m= (m_{p}+m_{e})-m_n$$
$$\Delta m=-9 \times 10^{-31}kg$$
Energy released $$=9 \times 10^{-31} \times (3\times 10^8)^2J$$
$$=\dfrac{9 \times 10^{-31} \times (3\times 10^8)^2}{1.6\times 10^{-13}}
$$
$$=0.50625\ \mathrm{MeV}$$
Question 4
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The energy spectrum of $$\beta- particles$$ [number $$N(E)$$ as a function of of $$\beta-energy\ E$$] emitted from a radioactive source is

A

B

C

D

Solution

The range of energy of $$\beta$$ particles is from zero to some maximum value. Graph (d) represents the variation
Question 5
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If the binding energy per nucleon in $$_{7}^{}\textrm{3}Li$$ and He nuclei are $$5.60 MeV$$ and $$7.06 MeV$$ respectively, then in the reaction $$p+_{3}^{}\textrm{7}Li\rightarrow _{2}^{}\textrm{4}$$ He, binding energy is

A
39.2 MeV

B
28.24 MeV

C
17.28 MeV

D
1.46 MeV

Solution

Since, $$p+_3Li^7 \rightarrow 2_2He^4$$
Thus, proton energy = BE of He atom $$-$$ BE of Li of atom.
or $$E_p=(2\times 4 \times 7.06 ) -(7\times 5.60)=17.28 MeV$$ where 4 is the no of nucleon of He and 7 is the no of nucleon of Li.
Question 6
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The binding energy per nucleon of deuteron $$(_{1}\mathrm{H}^{2})$$ and helium nucleus $$(_{2} \mathrm{He}^{4} )$$ is $$1.1 $$$$\mathrm{M}\mathrm{e}\mathrm{V}$$ and $$7$$$$\mathrm{M}\mathrm{e}\mathrm{V}$$ respectively. If two deuteron nuclei reacts to form a single helium nucleus, then the energy released is

A
$$13.9\mathrm{M}\mathrm{e}\mathrm{V}$$

B
$$26.9 \mathrm{M}\mathrm{e}\mathrm{V}$$

C
$$23.6 \mathrm{M}\mathrm{e}\mathrm{V}$$

D
$$19.2 \mathrm{M}\mathrm{e}\mathrm{V}$$

Solution

Energy required to break $$4$$ nucleons in the two deuterons $$=4\times 1.1\ MeV$$
Energy released in formation of $$4$$ nucleons in the single helium nucleus $$=4\times 7\ MeV$$
Hence, total energy released $$=(28-4.4)MeV=23.6\ MeV$$
Question 7
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The binding energy per nucleon for the parent nucleus is $$\mathrm{E}_{1}$$ and that for the daughter nuclei is $$\mathrm{E}_{2}$$.Then

A
$$\mathrm{E}_{2}=2\mathrm{E}_{1}$$

B
$$\mathrm{E}_{1}>\mathrm{E}_{2}$$

C
$$\mathrm{E}_{2}>\mathrm{E}_{1}$$

D
$$\mathrm{E}_{1}=2\mathrm{E}_{2}$$

Solution

After decay, the daughter nuclei will be more stable hence binding energy per nucleon will be more than that of their parent nucleus.
Question 8
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In gamma ray emission from a nucleus

A
both the neutron number and the proton number change

B
there is no change in the proton number and the neutron number

C
only the neutron number changes

D
only the proton number changes

Solution

In gamma decay, a nucleus changes from a higher energy state to a lower energy state through the emission of electromagnetic radiation(photons). The number of protons (and neutrons) in the nucleus does not change in this process, so the parent and daughter atoms are the same chemical element. In the gamma decay of a nucleus, the emitted photon and recoiling nucleus each have a well-defined energy after the decay. The characteristic energy is divided between only two particles.
Question 9
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The above is a plot of binding energy per nucleon $$\mathrm{E}_{\mathrm{b}}$$, against the nuclear mass $$\mathrm{M};\mathrm{A},\ \mathrm{B},\ \mathrm{C},\ \mathrm{D},\ \mathrm{E},\ \mathrm{F}$$ correspond to different nuclei. Consider four reactions :
(i) $$\mathrm{A}+\mathrm{B}\rightarrow \mathrm{C}+\epsilon$$
(ii) $$\mathrm{C}\rightarrow \mathrm{A}+\mathrm{B}+\epsilon$$
(iii) $$\mathrm{D}+\mathrm{E}\rightarrow \mathrm{F}+\epsilon$$
(iv) $$\mathrm{F}\rightarrow \mathrm{D}+\mathrm{E}+\epsilon$$
where $$\epsilon$$ is the energy released ? In which reaction is $$\epsilon$$ positive ?

A
(i) and (iv)

B
(i) and (iii)

C
(ii) and (iv)

D
(ii) and (iii).

Solution

The reactions for which product is/are more stable than reactant, energy & will be positive. (Energy will be released)
As B.E/nuclear decides stability
$$\therefore $$ Stability$$:C>B$$
$$C>A$$
$$D>F$$
and $$E>F$$
$$\left( i \right) A+B\longrightarrow C+\varepsilon $$
$$\left( iv \right) F\longrightarrow D+E+\varepsilon $$
are the reactions for which $$\varepsilon $$ will be positive
Question 10
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Directions For Questions

The mass of nucleus $$^{A}_{Z}X$$ is less than the sum of the masses of ($$\mathrm{A}-\mathrm{Z}$$) number of neutrons and $$\mathrm{Z}$$ number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. $$\mathrm{A}$$ heavy nucleus of mass $$\mathrm{M}$$ can break into two light nuclei of mass $$\mathrm{m}_{1}$$ and $$\mathrm{m}_{2}$$ only if $$(\mathrm{m}_{1}+\mathrm{m}_{2})<\mathrm{M}.$$ Also two light nuclei of masses $$\mathrm{m}_{3}$$ and $$\mathrm{m}_{4}$$ can undergo complete fusion and form a heavy nucleus of mass $$\mathrm{M}'$$ only if $$(\mathrm{m}_{3} +\mathrm{m}_{4})>\mathrm{M}'$$. The masses of some neutral atoms are given in the table above:

...view full instructions

The correct statement is

A
The nucleus $$^{6}_{3}Li$$ can emit an alpha particle

B
The nucleus $$^{210}_{84}Po$$ can emit a proton.

C
Deuteron and alpha particle can undergo complete fusion.

D
The nuclei $$^{70}_{30} Zn$$ and $$^{82}_{34}Se$$ can undergo complete fusion.

Solution

For fusion reaction to take place, sum of masses of lighter nuclei must be greater than mass of heavy nucleus formed.
For fusion reaction : $$^2_1H + ^4_2He\rightarrow ^6_3Li$$
Sum of mass of lighter nuclei $$M_l = 2.014102 + 4.002603 = 6.016705$$ u
Mass of $$^6_3Li$$ $$M_h = 6.015123$$ u
As $$M_l>M_h$$, thus above reaction is possible.