Nuclei Test

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Nuclei Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

As comparee to $$ ^{12}C $$ atom , $$ ^{14}C $$ atom has

A
two extra protons and two extra electrons

B
two extra protons but no extra electrons

C
two extra neutrons and no extra electron

D
two extra neutron and two extra electrons

Solution
Question 2
1 / -0

When number of nucleons in nuclei increases, the binding energy per nucleon:

A
increases continuously with mass number

B
decreases continuously with mass number

C
remain constant with mass number

D
first increases and then decreases with increase of mass number

Solution
Question 3
1 / -0

The mass and energy equivalent to $$1$$ a.m.u, respectively are :

A
$$1.67\times 10^{-27}\ g, 9.30\ MeV$$

B
$$1.67\times 10^{-27}\ kg, 930\ MeV$$

C
$$1.67\times 10^{-27}\ kg, 1\ MeV$$

D
$$1.67\times 10^{-34}\ kg, 1\ MeV$$

Solution
Question 4
1 / -0

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 the nucleus.

D
It is the sum of the kinetic energy of all the nucleons in the nucleus.

Solution
Question 5
1 / -0

A radioactive sample decays by $$63$$% of its initial value in $$10s$$. It would have decayed by $$50$$% of its initial value in

A
$$7s$$

B
$$14s$$

C
$$0.7s$$

D
$$1.4s$$

Solution

A radioactive sample decays by $$63$$% in $$t=\dfrac{1}{\lambda}$$
Hence $$\lambda=1/10$$
$$N=N_0e^{-\lambda t}$$
$$\dfrac{N_0}2=N_0e^{-0.1t}$$
$$0.1t=ln2$$
$$t=6.93s\approx7s$$
Question 6
1 / -0

During $$\beta$$-decay:

A
An atomic electron is ejected

B
An electron which is already present with in the nucleus is ejected

C
A neutron in the nucleus decays emitting an electron

D
A part of binding of the nucleus is converted into an electron

Solution

$$\beta-$$ decay transforms a neutron into a proton by the emission of an electron accompanied by an antineutrino.
Or, a proton is converted into a neutron by the emission of a positron with a neutrino in so-called positron emission.
Hence, option $$C$$ is correct.
Question 7
1 / -0

Consider one of fission reactions of $$_{ }^{ 235 }{ U }$$ by thermal neutrons $$_{ 92 }^{ 235 }{ U }+n\rightarrow _{ 38 }^{ 94 }{ Sr }+_{ 54 }^{ 140 }{ Xe }+2n$$. The fission fragments are however unstable and they undergo successive $$\beta$$- decay until $$_{ 38 }^{ 94 }{ Sr }$$ becomes $$_{ 40 }^{ 94 }{ Zr }$$ and $$_{ 54 }^{ 140 }{ Xe }$$ becomes $$_{ 58 }^{ 140 }{ Ce }$$. The energy released in this process is
[Given: $$m(_{ }^{ 235 }{ U })=235.439u$$; $$m(n)=1.00866u$$; $$m(_{ }^{ 94 }{ Zr }=93.9064u$$; $$m(_{ }^{ 140 }{ Ce })=139.9055u$$; $$1u=931MeV$$]

A
$$156MeV$$

B
$$208MeV$$

C
$$456MeV$$

D
Cannot be computed

Solution

$$_{92}^{235}U+n\longrightarrow _{38}^{94}Sr+_{54}^{140}Xe+2n$$

$$_{38}^{94}Sr\longrightarrow _{40}^{94}Zr+2e^-$$

$$_{54}^{140}Xe\longrightarrow _{58}^{140}Xe+4e^-$$

The complete fission reaction is

$$_{ 92 }^{ 235 }{ U }+n\rightarrow _{ 40 }^{ 94 }{ Zr }+_{ 58 }^{ 140 }{ Ce }+2n+6{ e }^{ -1 }$$
$$Q=\Delta mc^2$$
$$Q=\left[ m\left( _{ }^{ 235 }{ U } \right) -m\left( _{ }^{ 94 }{ Zr } \right) -m\left( _{ }^{ 140 }{ Ce } \right) -m(n) \right] { c }^{ 2 }=208MeV$$
Question 8
1 / -0

$$1.00kg$$ of $$_{ }^{ 235 }{ U }$$ undergoes fission process. If energy released per event is $$200MeV$$, then the total energy released is

A
$$5.12\times {10}^{24}MeV$$

B
$$6.02\times {10}^{23}MeV$$

C
$$5.12\times {10}^{26}MeV$$

D
$$6.02\times {10}^{26}MeV$$

Solution

The number of nuclei in $$1kg$$ $$_{ }^{ 235 }{ U }$$ is
$$N=\cfrac { { N }_{ A } }{ 235 } \times \left( 1\times { 10 }^{ 3 } \right) =2.56\times { 10 }^{ 24 }$$nuclei
Total energy released is
$$E=N\times 200MeV=5.12\times { 10 }^{ 26 }MeV$$
Question 9
1 / -0

Calculate the binding energy of a deutron atom, which consists of a proton and a neutron, given that the atomic mass of the deuteron is $$2.014102u$$

A
$$0.002388MeV$$

B
$$2.014102MeV$$

C
$$2.16490MeV$$

D
$$2.224MeV$$

Solution

atomic mass $$M(H)$$ of hydrogen and nuclear mass $$({M}_{n})$$ are
$$M(H)=1.007825u$$ and $${M}_{n}=1.008665u$$
Mass defect
$$\Delta m=\left[ M(H)+{ M }_{ n }-M(D) \right] ;M(D)=2.016490u-2.014102u=0.002388u$$
As $$1u$$ corresponds to $$931.494MeV$$ energy, therefore, mass defect corresponds to energy
$${ E }_{ b }=0.002388\times 931.5=2.224MeV\quad $$
Question 10
1 / -0

The radioactivity of a sample is $${R}_{1}$$ at a time $${T}_{1}$$ and $${R}_{2}$$ at a time $${T}_{2}$$. If the half-life of the specimen is $$T$$, the number of atoms that have disintegrated in the time $$({T}_{2}-{T}_{1})$$ is proportional to

A
$${R}_{1}{T}_{1}-{R}_{2}{T}_{2}$$

B
$${R}_{1}-{R}_{2}$$

C
$$\cfrac{({R}_{1}-{R}_{2})}{T}$$

D
$$({R}_{1}-{R}_{2})T$$

Solution

$${ R }_{ 1 }={ N }_{ 1 }\lambda$$
$${ R }_{ 2 }={ N }_{ 2 }{ \lambda }$$
$$R_1-R_2=(N_1-N_2)\lambda$$
Also
$$T=\cfrac { \log _{ e }{ 2 } }{ \lambda } \Rightarrow \lambda =\cfrac { \log _{ e }{ 2 } }{ T } $$
$${ R }_{ 1 }-{ R }_{ 2 }=\left( { N }_{ 1 }-{ N }_{ 2 } \right) \lambda$$
$$R_1-R_2=\left( { N }_{ 1 }-{ N }_{ 2 } \right) \cfrac { \log _{ e }{ 2 } }{ T }$$
$$ \Rightarrow \left( { N }_{ 1 }-{ N }_{ 2 } \right) \propto \left( { R }_{ 1 }-{ R }_{ 2 } \right) T$$

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Nuclei Test - 56

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Nuclei Test - 56

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Question 1

two extra protons and two extra electrons

two extra protons but no extra electrons

two extra neutrons and no extra electron

two extra neutron and two extra electrons

Solution

Question 2

increases continuously with mass number

decreases continuously with mass number

remain constant with mass number

first increases and then decreases with increase of mass number

Solution

Question 3

$$1.67\times 10^{-27}\ g, 9.30\ MeV$$

$$1.67\times 10^{-27}\ kg, 930\ MeV$$

$$1.67\times 10^{-27}\ kg, 1\ MeV$$

$$1.67\times 10^{-34}\ kg, 1\ MeV$$

Solution

Question 4

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 the nucleus.

It is the sum of the kinetic energy of all the nucleons in the nucleus.

Solution

Question 5

$$7s$$

$$14s$$

$$0.7s$$

$$1.4s$$

Solution

Question 6

An atomic electron is ejected

An electron which is already present with in the nucleus is ejected

A neutron in the nucleus decays emitting an electron

A part of binding of the nucleus is converted into an electron

Solution

Question 7

$$156MeV$$

$$208MeV$$

$$456MeV$$

Cannot be computed

Solution

Question 8

$$5.12\times {10}^{24}MeV$$

$$6.02\times {10}^{23}MeV$$

$$5.12\times {10}^{26}MeV$$

$$6.02\times {10}^{26}MeV$$

Solution

Question 9

$$0.002388MeV$$

$$2.014102MeV$$

$$2.16490MeV$$

$$2.224MeV$$

Solution

Question 10

$${R}_{1}{T}_{1}-{R}_{2}{T}_{2}$$

$${R}_{1}-{R}_{2}$$

$$\cfrac{({R}_{1}-{R}_{2})}{T}$$

$$({R}_{1}-{R}_{2})T$$

Solution

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