Nuclei Test

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Nuclei Test - 29

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Weekly Quiz Competition

Question 1
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The binding energies energy per nucleon for $$C^{12}$$ is $$7.68\ MeV$$ and that for $$C^{13}$$ is $$7.5\ MeV.$$ The energy required to remove a neutron from $$C^{13}$$ is

A
$$5.34\ MeV$$

B
$$5.5\ MeV$$

C
$$9.5\ MeV$$

D
$$9.34\ MeV$$

Solution

Total binding energy of $$C^{13}$$ is $$E_1=7.5\times 13=97.5\ MeV$$
Total binding energy of $$C^{12}$$ is $$E_2=7.68\times 12=92.16\ MeV$$
The energy required to remove a neutron is $$E_1-E_2=5.34\ MeV$$
Question 2
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Directions For Questions

The neutral ion $$\pi$$ is an unstable particle produced in high energy particle collisions. Its mass is about $$264$$ times the mass of electron. Note that its average life time is $$8.4\times10^{-17}$$ $$s$$ before decaying into two gamma ray photons. Using the energy mass relationship between rest mass and energy, one can find uncertainty in the mass.

...view full instructions

Find the uncertainty in mass.

A
$$1.4\times10^{-35}$$ $$kg$$

B
$$2.6\times10^{-11}$$ $$kg$$

C
$$2.6\times10^{-13}$$ $$kg$$

D
$$8\times10^{-14}$$ $$kg$$

Solution

$$\triangle E.\triangle t$$ = $$h/2\pi$$

$$\triangle E$$$$=\displaystyle\ \frac{h}{2\pi \triangle t}$$ = $$\displaystyle\ \frac

{6.62\times10^{-34}}{2\pi\times8.4\times10^{-17}}$$$$=1.24\times10^{-18}$$

$$\triangle mc^{2}$$ = $$\triangle E$$

Uncertainty in mass is $$\triangle m$$ $$=\displaystyle\ \frac{1.24\times10^{-18}}{9\times10^{16}}$$$$=1.38\times10^{-35}\approx 1.4\times10^{-35}$$ $$kg$$
Question 3
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Which of the following is a wrong description of binding energy of a nucleus?

A
It is the energy required to break a nucleus into its constituent nucleons

B
It is the energy made available when free nucleons combine to form a nucleus

C
It is the sum of the rest mass energies of its nucleons minus the rest mass energy of nucleus

D
It is the sum of the kinetic energy of all the nucelons in the nucleus

Solution

As, binding energy is the energy required to disassemble a whole system into separate parts. It is the energy that hold the nuclear together.
Question 4
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Find the binding energy of $$_{28}^{62}\textrm{Ni}$$, Given $$m_{H}$$ = $$1.008$$ $$u$$, $$m_{n}$$ = $$1.0087$$ $$u$$, $$_{28}^{62}\textrm{m}$$ = $$62.9237$$ $$u$$

A
$$545.3$$ $$MeV$$

B
$$595.3$$ $$MeV$$

C
$$645.3$$ $$MeV$$

D
$$695.3$$ $$MeV$$

Solution

Number of protons in $$_{28}^{62}Ni$$, $$p=28$$
Number of neutrons in $$_{28}^{62}Ni$$, $$n=28$$

Given: $$m_{H}=1.008u , m_{n}=1.0087u$$ ,
If $$_{28}^{62}m=61.9237u$$
Then mass defect,
$$\Delta m=28m_{H}+34m_{n}-_{28}^{34}m$$
$$\Delta m=28\times1.008+34\times1.0087-61.9237$$
$$\Delta m=0.5961u$$
Hence binding energy,
$$B.E.=\Delta m\times931$$
$$B.E.=0.5961\times930=554.3\ MeV$$
Question 5
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The fusion of two nuclide will require a temp of order of

A
$$10^{6}$$ $$K$$

B
$$10^{7}$$ $$K$$

C
$$10^{8}$$ $$K$$

D
$$10^{9}$$ $$K$$

Solution

Fusion requires temperatures about 100 million Kelvin (approximately six times hotter than the sun's core)
Question 6
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Directions For Questions

The neutral ion $$\pi$$ is an unstable particle produced in high energy particle collisions. Its mass is about $$264$$ times the mass of electron. Note that its average life time is $$8.4\times10^{-17}$$ $$s$$ before decaying into two gamma ray photons. Using the energy mass relationship between rest mass and energy, one can find uncertainty in the mass.

...view full instructions

Find the fraction of the mass.

A
$$5.28\times10^{-6}$$

B
$$5. 84\times10^{-7}$$

C
$$5.28\times10^{-8}$$

D
$$7.8\times10^{-9}$$

Solution

Given : $$\Delta t = 8.4\times 10^{-17}$$ s
As we know the relation $$\triangle E.\triangle t$$ $$=\dfrac{h}{2\pi}$$
$$\therefore $$ $$\triangle E$$$$=\displaystyle\ \frac{h}{2\pi \triangle t}$$ = $$\displaystyle\ \frac

{6.62\times10^{-34}}{2\pi\times8.4\times10^{-17}}=1.24\times10^{-18}$$ J
From mass-energy equivalance relation : $$\triangle mc^{2}$$ = $$\triangle E$$
Uncertainty in mass $$\triangle m$$ $$ = \dfrac{\Delta E}{c^2} =\displaystyle\ \frac{1.24\times10^{-18}}{(3\times10^{8})^2}=1.38\times10^{-35}\approx 1.4\times10^{-35}$$ $$kg$$
Mass of neutral pion $$m_{pion} = 264 m_{electron} = 264\times 9.1\times 10^{-31}$$ kg
$$\therefore$$ Fraction of mass $$\displaystyle\ \frac{\triangle m}{m_{pion}}$$ = $$\displaystyle\

\frac{1.38\times10^{-35}}{264\times9\times10^{-31}}=5.28\times10^{-8}$$
Question 7
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As the mass number $$A$$ increases, the binding energy per nucleon in a nucleus

A
increases

B
decreases.

C
remains the same.

D
varies in a way that depends on the actual value of $$A$$.

Solution

From binding energy curve,
The binding energy per nucleon varies with number of nucleons, A.
Question 8
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$$_{1}^{2}\textrm{H}$$ + $$_{4}^{9}\textrm{Be}$$ $$\rightarrow$$ $$X$$ + $$_{2}^{4}\textrm{He}$$ identify $$X$$

A
$$_{3}^{7}\textrm{Li}$$

B
$$_{3}^{6}\textrm{Li}$$

C
$$_{4}^{7}\textrm{Be}$$

D
$$2$$ $$_{2}^{3}\textrm{He}$$

Solution

The given eq. is:
$$_{1}^{2}H+_{4}^{9}Be\longrightarrow X+_{2}^{4}He$$

Let X is an atom with atomic no. x and atomic weight y i.e. $$_{x}^{y}X$$
For a balanced equation:
$$1+4=x+2$$
or $$x=3$$
and $$2+9=y+4$$
or $$y=7$$
which is best suited to $$_{3}^{7}Li$$ out of the given options .
Question 9
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Consider $$_{13}^{25}\textrm{Al}$$ $$\rightarrow$$ $$_{12}^{25}\textrm{Mg}$$ + $$_{+1}^{0}\textrm{e}$$+ $$v$$.
$$m_{Al}$$ = $$24.990423 \mu$$; $$m_{mg}$$= $$24.485839 \mu$$.
Find the Q value of reaction

A
$$3.3$$ $$Mev$$

B
$$1.3$$ $$Mev$$

C
$$2.3$$ $$MeV$$

D
$$5.3$$ $$MeV$$

Solution

$$Q$$ $$=[24.990432 \mu-24.98583 \mu-2m_{e}]c^{2}$$
$$=0.004593(931.5) - 1.102$$
$$=3.3$$ $$MeV$$
Question 10
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Which of the following reactions is impossible?

A
$$_{2}\textrm{He}^{4}$$ + $$_{4}\textrm{Be}^{9}$$ = $$_{0}\textrm{n}^{1}$$ +$$_{6}\textrm{C}^{12}$$

B
$$_{2}\textrm{He}^{4}$$ + $$_{7}\textrm{N}^{14}$$ = $$_{1}\textrm{H}^{1}$$ + $$_{8}\textrm{O}^{17}$$

C
4($$_{1}\textrm{H}^{1}$$) = $$_{2}\textrm{He}^{4}$$ + 2($$_{1}\textrm{e}^{0}$$)

D
$$_{3}\textrm{Li}^{4}$$ + $$_{1}\textrm{H}^{1}$$ = $$_{4}\textrm{Be}^{8}$$

Solution

In the nuclear reactions atomic mass of nucleons should remained conserved on either sides of reaction or mass of product nuclei should be less than the reacting ones.
A] In first reaction the total atomic mass of reactant nuclei i.e. $$He$$ and $$Be\ (9 + 4 = 13)$$ is equal to the mass of product nuclei i.e. C and neutron$$(12 + 1 = 13).$$
B] Also, in second reaction the total atomic mass of reactant nuclei i.e. $$He$$ and $$N (4 + 14 = 18) $$ is equal to the mass of product nuclei i.e. proton and oxygen$$(1 + 17 = 18)$$.
C] Similarly in third reaction mass is conserved between $$4$$ protons and alpha particle.
D] In this reaction, mass of lithium and proton is less than the isotope of $$Be$$ this is not possible since, the light nuclei combines to form the heavier one of equal mass or with some mass defect which is converted into energy. But in given reaction mass of product nucleus is large which is in contradiction.

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Nuclei Test - 29

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Nuclei Test - 29

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Question 1

$$5.34\ MeV$$

$$5.5\ MeV$$

$$9.5\ MeV$$

$$9.34\ MeV$$

Solution

Question 2

$$1.4\times10^{-35}$$ $$kg$$

$$2.6\times10^{-11}$$ $$kg$$

$$2.6\times10^{-13}$$ $$kg$$

$$8\times10^{-14}$$ $$kg$$

Solution

Question 3

It is the energy required to break a nucleus into its constituent nucleons

It is the energy made available when free nucleons combine to form a nucleus

It is the sum of the rest mass energies of its nucleons minus the rest mass energy of nucleus

It is the sum of the kinetic energy of all the nucelons in the nucleus

Solution

Question 4

$$545.3$$ $$MeV$$

$$595.3$$ $$MeV$$

$$645.3$$ $$MeV$$

$$695.3$$ $$MeV$$

Solution

Question 5

$$10^{6}$$ $$K$$

$$10^{7}$$ $$K$$

$$10^{8}$$ $$K$$

$$10^{9}$$ $$K$$

Solution

Question 6

$$5.28\times10^{-6}$$

$$5. 84\times10^{-7}$$

$$5.28\times10^{-8}$$

$$7.8\times10^{-9}$$

Solution

Question 7

increases

decreases.

remains the same.

varies in a way that depends on the actual value of $$A$$.

Solution

Question 8

$$_{3}^{7}\textrm{Li}$$

$$_{3}^{6}\textrm{Li}$$

$$_{4}^{7}\textrm{Be}$$

$$2$$ $$_{2}^{3}\textrm{He}$$

Solution

Question 9

$$3.3$$ $$Mev$$

$$1.3$$ $$Mev$$

$$2.3$$ $$MeV$$

$$5.3$$ $$MeV$$

Solution

Question 10

$$_{2}\textrm{He}^{4}$$ + $$_{4}\textrm{Be}^{9}$$ = $$_{0}\textrm{n}^{1}$$ +$$_{6}\textrm{C}^{12}$$

$$_{2}\textrm{He}^{4}$$ + $$_{7}\textrm{N}^{14}$$ = $$_{1}\textrm{H}^{1}$$ + $$_{8}\textrm{O}^{17}$$

4($$_{1}\textrm{H}^{1}$$) = $$_{2}\textrm{He}^{4}$$ + 2($$_{1}\textrm{e}^{0}$$)

$$_{3}\textrm{Li}^{4}$$ + $$_{1}\textrm{H}^{1}$$ = $$_{4}\textrm{Be}^{8}$$

Solution

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