Nuclei Test

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Nuclei Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

Binding energy per nucleon versus 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
$${\rm{Y}} \to {\rm{2Z}}$$

B
$${\rm{W}} \to {\rm{X + Z}}$$

C
$${\rm{W}} \to {\rm{2Y}}$$

D
$${\rm{X}} \to {\rm{Y + Z}}$$

Solution

If it is mass A
$$Y\rightarrow 2Z$$
Reactant $$R=60\times8.5=10meV$$
Product $$P=2\times30\times5=300meV$$
$$\Delta E=-210meV$$
Therefore Endothermic
If it was C
$$W\rightarrow 2Y$$
$$R=120\times75=900meV$$
$$P=2\times60\times8.5=1020meV$$
$$\Delta E=120meV$$
Therefore Exothermic
If it was D
$$X\rightarrow Y+Z$$
$$R=90\times8=720meV\\P=60\times8.5+30\times5=66.0meV$$
$$\Delta E=-60meV$$
Therefore Endothermic
Question 2
1 / -0

A stationary $$U^{238}$$ nucleus undergoes $$\alpha$$- decay. If the kinetic energy of product nucleus is $$E$$, the total energy released in the process is-

A
$$E$$

B
$$58.5E$$

C
$$59.5E$$

D
$$60.5E$$

Solution

mass of alpha particle $$m_a=4m$$, where $$m$$ is mass of a proton.
mass of Uranium nucleus is $$m_u=238m$$ after emission the residue will have mass $$m_r=238-4=234m$$

We know $$KE=momentum^2/2m$$
so for residue nucleus, it should be like this $$E=p^2_r/2m_r$$

or momentum of residue nucleus is $$p_r=\sqrt[2]{2m_rE}=\sqrt[2]{468mE}$$
But during emission the alpha particle should get same momentum in opposite direction to make net momentum as zero
before and after emission so it should also be $$p_a=\sqrt[2]{468mE}$$

If kinetic energy of alpha particle is $$E_a$$ then $$E_a=p^2_a/2m_a=\dfrac{468mE}{2\times 4m}=58.5E$$

So net energy released in the emission is the sum of above two i.e. $$59.5E$$
Question 3
1 / -0

The binding energy per nucleon of deuteron is $$1.2MeV$$ and that of the helium atom is $$7.1MeV$$. What is the energy released, if two deuteron atoms combine to form a single helium atom?

A
17.8MeV

B
19.5MeV

C
21,2MeV

D
23.6MeV

Solution

$$\begin{array}{c}\text { Given - } * B E \text { per nucleon of deuteron }=1.2 \mathrm{MeV} \\\text { and of Helium }=7.1\mathrm{MeV}\\{ }_{1}^{2} \mathrm{H}+{}_{1}^{2}\mathrm{H}\longrightarrow_{4}^{2}\mathrm{He}\end{array}$$

$$\begin{array}{l}\Rightarrow \text { Energy released = BE of Helium -2.BE of Deutron. } \\\text { BE }=B E \text { ber nucleons } X \text { Num of nucleons. } \\\begin{aligned}\Rightarrow E &=[4 \times 7.1]-2\times[1.2 \times 2] \text { Mev } \\E &=23.6\mathrm{MeV}\end{aligned}\end{array}$$
Question 4
1 / -0

The number of atoms of the $$He$$ in $$104\ amu$$ is:

A
$$3.1 \times 10^{25}$$

B
$$6.2 \times 10^{25}$$

C
$$26$$

D
$$206$$

Solution

Since the mass of one atom of He = 4 a.m.u
Therefore the number of atoms of the He in 104 amu is, $$\dfrac { 104 }{ 4 } =26$$.
Question 5
1 / -0

What would be the energy required to dissociate completely $$1$$ g of $$Ca \ (40)$$ into its constituent particles?
Given: Mass of proton = 1.007277 amu,
Mass of neutron = 1.00866 amu,
Mass of Ca-40= 39.97545 amu
(take amu = 931 MeV)

A
$$4.813\times 10^{24}$$MeV

B
$$4.813\times 10^{24}$$ eV

C
$$4.813\times 10^{23}$$

D
None of the above

Solution

Given,
$$m_p=1.007277 amu$$ mass of proton
$$m_n=1.00866amu$$ mass of neutron
$$m_c(ca-40)=39.97545 amu$$
and $$1amu=931 MeV/c^2$$
$$Z=20$$ No of proton for calcium
$$N=20$$ no. of neutron

The energy required to dissociate completely is called binding energy.
$$B.E=(Zm_p+Nm_n-m_c)c^2$$ . . . .(1)

$$B.E=(20\times 1.007277+20\times 1.00866-39.97545)c^2$$
$$B.E=0.34359\times 931MeV$$

$$B.E=319.6MeV$$ This is for 1 atom of calcium-40

No. of atom in $$1g$$ of Ca-40 is,
$$N=\dfrac{1}{40}\times 6.6022\times 10^{23}$$

$$N=1.65055\times 10^{22}$$ atom.
The total energy required is,
$$E=N(B.E)$$
$$E=1.65055\times 10^ {22}\times319.6MeV $$
$$E=4.813\times 10^{24}MeV$$
The correct option is A.
Question 6
1 / -0

The binding energy of deutron is $$2.2 \space MeV$$ and that of $$^4_2 He$$ is $$28 \space MeV$$. If two deutrons are fused to form one $$^4_2 He$$ then the energy released is:

A
$$25.8 \space MeV$$

B
$$23.6 \space MeV$$

C
$$19.2 \space MeV$$

D
$$30.2 \space MeV$$

Solution

$$_1{D^2}{ \to _2}H{e^4}$$
Energy released $$=28 - 2 \times 2.2 = 23.6\,MeV$$
Hence,
option $$(B)$$ is correct answer.
Question 7
1 / -0

If $$x \ g$$ of $$A$$ (atomic mass $$50$$) contains $$n$$ atoms, how many atoms are there in $$20 \ x \ g \ B$$ (atomic weight $$100$$)

A
$$n$$

B
$$10 n$$

C
$$20 n$$

D
$$\dfrac{n}{10}$$

Solution
Question 8
1 / -0

The binding energy per nucleon of $$^7_3 Li$$ and $$^4_2 He$$ nuclei are 5.60 MeV and 7.06 Me V, respectively. In the nuclear reaction $$^7_3 Li + ^1_1 He \longrightarrow ^4_2 H + ^4_2 He + Q$$ the value of energy Q released is

A
-2.4 MeV

B
8.4 MeV

C
17.3 MeV

D
19.6 MeV

Solution

The binding energy for $$_1H^1$$ is around zero and also not give in the question so we can ignore it,
$$Q=2(4\times 7.06)-7\times (5.60)$$
$$=(56.48-39.2)MeV$$
$$=17.28\ MeV$$
$$=17.3\ MeV$$
Question 9
1 / -0

Neutron decay in free space is given as follows:
$$_{e}n^{1} \rightarrow_{0}H^{1}+_{-o}e^{0}+[]$$
Then the parenthes is represents a

A
Neutrion

B
Photon

C
Antineutrino

D
Gravition

Solution

$$_{e}{n}^{1}\rightarrow_{0}H^{1}+_{-1}e^{0}+\text { some particles }$$
$$\begin{array}{l}\text { The given decay is similar to } \beta^{-} \text {decay. } \\\text { To maintain the momentum, angular momentum } \\\text { spin moment. The particle released is antineutrino. }\end{array}$$
Question 10
1 / -0

A slow neutron strikes a nucleus of $$^{235}_{92} {U}$$ splitting it into lighter nuclei of $$^{141}_{56} {Ba}$$ and $$^{92}_{36} {Kr}$$ along with three neutrons. The energy released in this reaction is :
(The masses of Uranium, Barium and Krypton in this reaction are $${235.043933}$$ a.m.u, $${140.917700}$$ a.m.u and $${91.895400}$$ a.m.u respectively. The mass of a neutron is $$1.008665$$ a.m.u)

A
$$198.9$$ MeV

B
$$156.9$$ MeV

C
$$186.9$$ MeV

D
$$209.8$$ MeV

Solution

$${}_{92}^{235}U + {}_0^1n \to {}_{56}^{141}Ba + {}_{36}^{92}Kr + 3({}_0^1n) + Q$$

Q is Energy released Q = ( Total mass of reactants – total mass ) $$ \times {C^2}$$

$$ = \left[ {Mass\left( {{}_{92}^{235}U} \right) + mass\left( {{}_0^1n} \right)} \right] - \left[ {mass{}_{56}^{141}Ba + mass{}_{36}^{92}Kr + mass3{}_0^1n} \right]{c^2}$$

$$ = \left( {235 \cdot 043933 - 140.917700 - 91.8954 - 2 \times 1.008665} \right)u \times {c^2}$$

$$ = 198.9MeV$$

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Nuclei Test - 48

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Nuclei Test - 48

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Question 1

$${\rm{Y}} \to {\rm{2Z}}$$

$${\rm{W}} \to {\rm{X + Z}}$$

$${\rm{W}} \to {\rm{2Y}}$$

$${\rm{X}} \to {\rm{Y + Z}}$$

Solution

Question 2

$$E$$

$$58.5E$$

$$59.5E$$

$$60.5E$$

Solution

Question 3

17.8MeV

19.5MeV

21,2MeV

23.6MeV

Solution

Question 4

$$3.1 \times 10^{25}$$

$$6.2 \times 10^{25}$$

$$26$$

$$206$$

Solution

Question 5

$$4.813\times 10^{24}$$MeV

$$4.813\times 10^{24}$$ eV

$$4.813\times 10^{23}$$

None of the above

Solution

Question 6

$$25.8 \space MeV$$

$$23.6 \space MeV$$

$$19.2 \space MeV$$

$$30.2 \space MeV$$

Solution

Question 7

$$n$$

$$10 n$$

$$20 n$$

$$\dfrac{n}{10}$$

Solution

Question 8

-2.4 MeV

8.4 MeV

17.3 MeV

19.6 MeV

Solution

Question 9

Neutrion

Photon

Antineutrino

Gravition

Solution

Question 10

$$198.9$$ MeV

$$156.9$$ MeV

$$186.9$$ MeV

$$209.8$$ MeV

Solution

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