Nuclei Test

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Nuclei Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

Binding energy per nucleon vs. mass number curve for nuclei is shown in fig. W, X, Y and Z are four nuclei indicated on the curve. The process that would release energy is

A
$$Y\rightarrow 2Z$$

B
$$W\rightarrow X+Z$$

C
$$W\rightarrow 2Y$$

D
$$X\rightarrow Y+Z$$

Solution

Binding energy of $$W$$, $$|E_W| = 120\times7.5=900$$ MeV
Binding energy of $$Y$$, $$|E_Y| = 30\times8.5=255$$ MeV
In process : $$W\rightarrow 2Y$$
$$Q$$ value, $$Q = |E_W| - |E_Y| = 900-2(255) =390$$ MeV
As $$Q$$ value is positive, thus energy is released in process $$W\rightarrow 2Y$$
Question 2
1 / -0

Biding energy is equal to: (m is the mass defect in amu)

A
$$\Delta m \times 931 MeV$$

B
$$\Delta m \times 831 MeV$$

C
$$\Delta m \times 1031 MeV$$

D
$$\Delta m \times 731 MeV$$

Solution

From mass energy equivalence Binding energy,
$$E=\triangle m{ c }^{ 2 }\\ \because { c }^{ 2 }=931MeV\\ E=\triangle m \times 931MeV$$
Question 3
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The binding energies per nucleon for a deuteron and an -particle are $${ x }_{ 1} ,{ x}_{ 2 }$$ respectively. What will be the energy Q released in the reaction $$_{ 1 }^{ }{ { H }_{ }^{ 2 } }+_{ 1 }^{ }{ { H }_{ }^{ 2 } }\rightarrow _{ 2 }^{ }{ { He }_{ }^{ 4 } }+Q$$

A
$$4\left( { x }_{ 1 }+{ x }_{ 2 } \right)$$

B
$$4\left( { x }_{ 2 }-{ x }_{ 1 } \right)$$

C
$$2\left( { x }_{ 1 }+{ x }_{ 2 } \right)$$

D
$$2\left( { x }_{ 2 }-{ x }_{ 1 } \right)$$

Solution
Question 4
1 / -0

The rest energy of an electron is $$0.511 MeV.$$ The electron is accelerated from rest to a velocity $$0.5 c$$. The change in its energy will be

A
$$0.026 MeV$$

B
$$0.051 MeV$$

C
$$0.07 MeV$$

D
$$0.105 MV$$

Solution

$$\displaystyle mc^2=\dfrac{m_0}{\sqrt{1-\dfrac{v^2}{c^2}}}c^2=\frac{m_0}{\sqrt{1-\dfrac{(0.5c)^2}{c^2}}}c^2$$$$\displaystyle =\frac{m_0}{\sqrt{1-0.25}}c^2=\frac{m_0}{\sqrt{0.75}}c^2=1.15m_0c^2$$

Change in energy $$= 1.15\:m_0c^2- m_0c^2$$
$$=0.15\times 9.1 \times 10^{-31}\times(3\times 10^8)^2$$
$$=12.285\times 10^{-15}J $$
$$\displaystyle =\frac{12.285\times10^{-15}}{1.6\times10^{-13} }MeV$$
$$= 0.07678\,MeV $$
Question 5
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Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which of these is most promising based on Lawson criterion?

A
Deuteron density=2.0 x $${ 10 }^{ 12 }{ cm }^{ -3 }$$ ,confinement time=5.0 x $${ 10 }^{ -3 }$$s

B
Deuteron density=8.0 x $${ 10 }^{ 14 }{ cm }^{ -3 }$$ ,confinement time=9.0 x $${ 10 }^{ -1 }$$s

C
Deuteron density=4.0 x $${ 10 }^{ 23 }{ cm }^{ -3 }$$ ,confinement time=1.0 x $${ 10 }^{ -11 }$$s

D
Deuteron density=1.0 x $${ 10 }^{ 24 }{ cm }^{ -3 }$$ ,confinement time=4.0 x $${ 10 }^{ -12 }$$s

Solution

According to Lawson, deuteron density (n) and confinement tome $$t_0$$ must satisfy the criterion
$$nt_0 > 5\times 10^{14} cm^{-3}s$$
This condition is satisfied only by choice(b) for which $$nt_0=(8.0\times 10^{14})\times (9.0\times 10^{-1})=7.2\times 10^{14}cm^{-3}s$$. Hence the correct choice is $$(b)$$.
Question 6
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Assume that two deutron nuclei in the core of fusion reactor at temperature $$T$$ are moving towards each other,each with kinetic energy $$1.5kT$$, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature $$T$$ required for them to reach a separation of $$4\times { 10 }^{ -15 }$$m is in the range

A
$$1.0\times { 10 }^{ 9 }K$$ < $$T$$<$$2.0\times { 10 }^{ 9 }K$$

B
$$2.0\times { 10 }^{ 9 }K $$< $$T$$<$$3.0\times { 10 }^{ 9 }K$$

C
$$3.0\times { 10 }^{ 9 }$$K < $$T$$<$$4.0\times { 10 }^{ 9 }K$$

D
$$4.0\times { 10 }^{ 9 }K$$ < $$T$$<$$5.0\times { 10 }^{ 9 }K$$

Solution

Applying conservation of mechanical energy, we get:
Loss of kinetic energy of two deuteron nuclei $$=$$ Gain in their potential energy.
$$\Rightarrow 2\times 1.5kT=\displaystyle\frac { 1 }{ 4\pi \varepsilon _{ 0 } } \frac { e\times e }{ r } $$
$$\Rightarrow 2\times 1.5\left( 8.6\times { 10 }^{ -5 }\frac { eV }{ K } \right) \times T=\displaystyle \frac { \left( 1.44\times { 10 }^{ -9 }eVm \right) }{ 4\times { 10 }^{ -15 }m } $$
$$\Rightarrow T=\displaystyle \frac { 1.44\times { 10 }^{ -9 } }{ 2\times 1.5\times 8.6\times { 10 }^{ -15 }\times 4\times { 10 }^{ -15 } } $$$$=0.0139\times { 10 }^{ 11 }=1.4\times { 10 }^{ 9 }K$$
Question 7
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Directions For Questions

Some of the heavy nuclei decay into lighter nuclei by emitting a-particle and acquire stability. The principles of conservation of atomic number, mass number, linear momentum and energy (including rest mass energy) are valid for these decays.
A stationary nucleus $$^{226}_{88}Ra$$ (ground state) decays into the nucleus $$^{222}_{86}Rn$$ (ground state) by emitting an $$\alpha - particle$$.
Given that $$m(^{226}_{88}Ra)=226.02540 u$$. $$m(^4_2He)=4.00260 u$$. $$m(^{222}_{86}Rn)=222.01750 u$$.

...view full instructions

The Q-value for the $$\alpha-decay$$ of $$^{226}_{88}Ra$$ is (approximately)

A
4.93 MeV

B
2.46 MeV

C
9.8 MeV

D
14.7 MeV

Solution

$$ \begin{array}{c} _{88}^{226} Ra \longrightarrow{ }_{2}^{4} He^{2+}+{ }_{86}^{222} Rn \end{array} $$
$$ 226.02540 \mu \quad 4.00260 \mu \quad 222.01750 \mu $$
$$ \begin{aligned} \Delta m=& 0.0053 \\ Q \text { -value } &=\Delta m \times 931.5 \mathrm{MeV} \\ &=0.0053 \times 931.5 \\ &=4.93695 \end{aligned} $$
Question 8
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Which of the following nuclear reactions is not possible?

A
$$ _{ 6 }^{ 12 }{ C+ }_{ 6 }^{ 12 }{ C }\longrightarrow _{ 10 }^{ 20 }{ Ne+ }_{ 2 }^{ 4 }{ He }$$

B
$$_{ 4 }^{ 9 }{ Be+ }_{ 1 }^{ 1 }{ H }\longrightarrow _{ 3 }^{ 6 }{ Li+ }_{ 2 }^{ 4 }{ He }$$

C
$$_{ 5 }^{ 11 }{ Be+ }_{ 1 }^{ 1 }{ H }\longrightarrow _{ 4 }^{ 9 }{ Be+ }_{ 2 }^{ 4 }{ He }$$

D
$$_{ 3 }^{ 7 }{ Li+ }_{ 2 }^{ 4 }{ He }\longrightarrow _{ 1 }^{ 1 }{ H+ }_{ 4 }^{ 10 }{ B }$$

Solution

The following laws hold true in any reaction
1) Conservation of mass (nucleons)
2) conservation of charge
Checking for all the reactions
$$\Sigma A_{products} = \Sigma A_{reactants}$$
For option C: $$ 9 +4 \ne 11+1$$
Hence, option (c) not possible.
Question 9
1 / -0

The velocity of a body of rest mass $$m_o$$ is $$\dfrac{\sqrt 3}{2} c$$ (where c is the velocity of light in vacuum). The mass of this body is :

A
$$ \left ( \dfrac{\sqrt3}{2} \right )m_o$$

B
$$\left ( \dfrac{1}{2} \right )m_o$$

C
$$3m_o$$

D
$$2m_o$$

Solution

By special theory of relativity
$$m' = \gamma m= \cfrac{m_0}{\sqrt{1 -\cfrac{v^2}{c^2}}}$$
$$m' = \cfrac{m_0}{\sqrt{1 -\cfrac{3}{4}}} =\cfrac{m_0}{(1/2)}= 2m_0$$
Question 10
1 / -0

Calculate the binding energy of $$_{3}^{6}\textrm{Li}$$ assuming the mass of $$_{3}^{6}\textrm{Li}$$ atom as $$6.01512$$ amu:

A
$$20.42 MeV$$

B
$$30.42 MeV$$

C
$$23.08 MeV$$

D
$$32.78 MeV$$

Solution

Mass defect is
$$\Delta m = [Zm_p + (A - Z)m_n-M]$$
$$m_p =$$ mass of the proton $$= 1.007277 amu$$
$$m_n =$$ mass of the neutron $$= 1.008655 amu$$
Mass of the bound $$_{3}^{6}\textrm{Li}$$, lithium has three protons and three neutrons, since $$Z = 3, A = 6$$ and number of neutrons $$=A -Z=6-3=3)$$

$$\therefore \Delta m = [3 \times 1.007277 + 3(1.008655] - 6.01512$$
$$=[3.02183 + 3.02596] - 6.01512$$
$$= 6.047795 - 6.01512 = 0.032675 amu.$$
But the energy equivalent of $$1 amu = 931 MeV$$
Therefore, the binding energy which is equal to the energy equivalent of mass defect in amu is
$$E_b =\Delta m \times 931 MeV$$
$$= 0.032675 \times 931 = 30.42 MeV$$

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Nuclei Test - 63

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Nuclei Test - 63

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Question 1

$$Y\rightarrow 2Z$$

$$W\rightarrow X+Z$$

$$W\rightarrow 2Y$$

$$X\rightarrow Y+Z$$

Solution

Question 2

$$\Delta m \times 931 MeV$$

$$\Delta m \times 831 MeV$$

$$\Delta m \times 1031 MeV$$

$$\Delta m \times 731 MeV$$

Solution

Question 3

$$4\left( { x }_{ 1 }+{ x }_{ 2 } \right)$$

$$4\left( { x }_{ 2 }-{ x }_{ 1 } \right)$$

$$2\left( { x }_{ 1 }+{ x }_{ 2 } \right)$$

$$2\left( { x }_{ 2 }-{ x }_{ 1 } \right)$$

Solution

Question 4

$$0.026 MeV$$

$$0.051 MeV$$

$$0.07 MeV$$

$$0.105 MV$$

Solution

Question 5

Deuteron density=2.0 x $${ 10 }^{ 12 }{ cm }^{ -3 }$$ ,confinement time=5.0 x $${ 10 }^{ -3 }$$s

Deuteron density=8.0 x $${ 10 }^{ 14 }{ cm }^{ -3 }$$ ,confinement time=9.0 x $${ 10 }^{ -1 }$$s

Deuteron density=4.0 x $${ 10 }^{ 23 }{ cm }^{ -3 }$$ ,confinement time=1.0 x $${ 10 }^{ -11 }$$s

Deuteron density=1.0 x $${ 10 }^{ 24 }{ cm }^{ -3 }$$ ,confinement time=4.0 x $${ 10 }^{ -12 }$$s

Solution

Question 6

$$1.0\times { 10 }^{ 9 }K$$ < $$T$$<$$2.0\times { 10 }^{ 9 }K$$

$$2.0\times { 10 }^{ 9 }K $$< $$T$$<$$3.0\times { 10 }^{ 9 }K$$

$$3.0\times { 10 }^{ 9 }$$K < $$T$$<$$4.0\times { 10 }^{ 9 }K$$

$$4.0\times { 10 }^{ 9 }K$$ < $$T$$<$$5.0\times { 10 }^{ 9 }K$$

Solution

Question 7

4.93 MeV

2.46 MeV

9.8 MeV

14.7 MeV

Solution

Question 8

$$ _{ 6 }^{ 12 }{ C+ }_{ 6 }^{ 12 }{ C }\longrightarrow _{ 10 }^{ 20 }{ Ne+ }_{ 2 }^{ 4 }{ He }$$

$$_{ 4 }^{ 9 }{ Be+ }_{ 1 }^{ 1 }{ H }\longrightarrow _{ 3 }^{ 6 }{ Li+ }_{ 2 }^{ 4 }{ He }$$

$$_{ 5 }^{ 11 }{ Be+ }_{ 1 }^{ 1 }{ H }\longrightarrow _{ 4 }^{ 9 }{ Be+ }_{ 2 }^{ 4 }{ He }$$

$$_{ 3 }^{ 7 }{ Li+ }_{ 2 }^{ 4 }{ He }\longrightarrow _{ 1 }^{ 1 }{ H+ }_{ 4 }^{ 10 }{ B }$$

Solution

Question 9

$$ \left ( \dfrac{\sqrt3}{2} \right )m_o$$

$$\left ( \dfrac{1}{2} \right )m_o$$

$$3m_o$$

$$2m_o$$

Solution

Question 10

$$20.42 MeV$$

$$30.42 MeV$$

$$23.08 MeV$$

$$32.78 MeV$$

Solution

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