Nuclei Test

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Nuclei Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

In a fission reaction $$^{236}_{92}U \rightarrow ^{117}X+^{117}Y+n+n$$. The binding energy per nucleon of $$X$$ and $$Y$$ are $$8.5 MeV$$ whereas of $$^{236}U$$ is $$7.6 MeV$$. The total energy liberated will be about

A
$$400 MeV$$

B
$$200 MeV$$

C
$$300 MeV$$

D
$$200 keV$$

Solution

$$\begin{array}{l}\text { Total no. of nucleus in " } X \text { " }\\\text { and "Y" would be } \\\qquad \begin{aligned}117+117=234 &\\\text { Total B.E of } X \text { and } Y=8.5 \times 234 \\=1989\text { MeV }\end{aligned}\end{array}$$

$$\begin{array}{rl}^{236}U \ have \ B E=7.6 \times 236 \\& =1793.6\mathrm{MeV} \\\triangle E & =1989-1793.6 \\\Delta E & =195.4\mathrm{MeV} \\\triangle E & \approx 200 \mathrm{Mev}\end{array}$$
$$\text{Approximately}$$
Question 2
1 / -0

Consider the nuclear reaction $$X^{200}\rightarrow A^{110}+B^{90}$$ if the binding energy per nucleon for X, A and B is 7.4 MeV, 8.2 MeV and 8.2 Mev respectively, what is the energy released?

A
200 MeV

B
160 MeV

C
110 MeV

D
90 MeV

Solution

Total energy of $$X=2\omega\times 7.4\ MeV$$
Total energy of $$A=110\times 8.2\ MeV$$
Total energy of $$B=90\times 78.2\ MMeV$$
So, $$2\omega\times 7.4=(110\times 8.2)+(90\times 8.2)$$
Hence $$-$$energy $$=-2\omega\times 7.4+(110\times 8.2)+(90\times 8.2)\\=160\ MeV$$
Question 3
1 / -0

Electrons are bombarded to excite hydrogen atoms and six spectral lines are observed. If $$E_{g}$$ is the ground state energy of hydrogen, the minimum energy the bombarding electrons should possess is

A
$$8E_{g}/9$$

B
$$15E_{g}/16$$

C
$$35E_{g}/36$$

D
$$48E_{g}/49$$

Solution
Question 4
1 / -0

A $$1.0\ MeV$$ neutron emitted in a fusion loses one half of its kinetic energy in each collision with a moderator molecule. Number of collision it must be more to reach thermal energy will be $$\left(Take, \dfrac {3}{2}k_{B}T=0.040\ eV\right)$$

A
$$5$$

B
$$25$$

C
$$125$$

D
$$625$$

Solution

$$\begin{array}{l}\text { Let "n" be required number } \\\text { of collision } \\\text { Given } \quad \frac{3}{2} k_{B} T=0.04 \\(0.04) n=1 \\\qquad n=25\end{array}$$
Question 5
1 / -0

Complete the equation for the following fission process $$_{92}U^{235}+_{0}n^{1}\rightarrow _{38}Sr^{90}+...$$

A
$$_{54}Xe^{143}+3_{0}n^{1}$$

B
$$_{54}Xe^{145}$$

C
$$_{57}Xe^{142}$$

D
$$_{54}Xe^{142}+3_{0}n$$

Solution

Sum of mass number in $$reactant $$ side is $$235+1=236$$
and Sum of atomic number in $$reactant $$ side is $$92+0=92$$
so the product side sum of mass number should also be $$236$$ and that of atomic number should be $$92$$
so $$_{54}Xe^{143} +3 _0n^1$$ is
suitable because $$143+3\times 1 +90=236$$ and $$38+54=92$$
Question 6
1 / -0

The reactor produced a power of $$10^{10}\ MW$$ with $$50$$% efficiency. Then the number of deutrons required to run the reactor for one year will be approximately:
$$m\left( _{ 1 }^{ }{ { H }_{ }^{ 2 } } \right) =2.014\ u,m\left( p \right) =1.007\ u,m\left( n \right) =1.008\ u,m\left( _{ 2 }^{ }{ { H }e_{ }^{ 4 } } \right) =4.001\ u$$

A
$$5\times 10^{35}$$

B
$$5\times 10^{32}$$

C
$$5\times 10^{38}$$

D
$$5\times 10^{40}$$
Question 7
1 / -0

The energy of the reaction $$Li ^ { 7 } + p \longrightarrow _2He^4$$ is (the binding energy per nucleon in $$Li^7$$ and $$He^4$$ nuclei are $$5.60$$ and $$7.06\ MeV$$ respectively.)

A
$$17.3\ MeV$$

B
$$1.73\ MeV$$

C
$$1.46\ MeV$$

D
$$Depends\ on\ the\ binding\ energy\ of\ proton.$$

Solution

$$\begin{aligned}\text { Energy of } r \times n &=B \cdot E_{p}-B \cdot E_{R} \\&=7.6 \times 4 \times 2-5.6 \times 7 \\&={17.3 \mathrm{ Mev}}\end{aligned}$$
Question 8
1 / -0

To start an endoergic reaction, a stationary nucleus $$(x)$$ in bombarded by a neutron $$(n)$$ to produce nuclei $$Y$$ and $$Z$$ in the final state. Choose the correct option from the following statements related to this reaction.

A
Kinetic energy of the bombarding neutron must be equal to the magnitude of the $$Q-value$$ of the reaction

B
Kinetic energy of the bombarding neutron must be less than the ,magnetic

C
Kinetic energy of the bombarding neutron must be greater than then magnetic

D
In this type of reaction, principle of conservation of energy is not valid
Question 9
1 / -0

Atomic mass of $$_{ 6 }^{ 13 }{ C }$$ is $$13.00335$$ amu and its mass number is $$13.0$$. If amu $$=931$$MeV, the binding energy of the neutrons present in the nucleus is

A
$$0.24$$ MeV

B
$$1.44$$MeV

C
$$1.68$$MeV

D
$$3.12$$MeV

Solution

binding energy is the product of mass defect and c square .
now 13.00335 - 13 = 0.00335u.
=>B.E = 931$$\times0.00335$$ = 3.12 Mev
Question 10
1 / -0

Nucleus of an elements contains 9 protons. It's valency would be:

A
1

B
2

C
3

D
5

Solution

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Nuclei Test - 51

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Nuclei Test - 51

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Question 1

$$400 MeV$$

$$200 MeV$$

$$300 MeV$$

$$200 keV$$

Solution

Question 2

200 MeV

160 MeV

110 MeV

90 MeV

Solution

Question 3

$$8E_{g}/9$$

$$15E_{g}/16$$

$$35E_{g}/36$$

$$48E_{g}/49$$

Solution

Question 4

$$5$$

$$25$$

$$125$$

$$625$$

Solution

Question 5

$$_{54}Xe^{143}+3_{0}n^{1}$$

$$_{54}Xe^{145}$$

$$_{57}Xe^{142}$$

$$_{54}Xe^{142}+3_{0}n$$

Solution

Question 6

$$5\times 10^{35}$$

$$5\times 10^{32}$$

$$5\times 10^{38}$$

$$5\times 10^{40}$$

Question 7

$$17.3\ MeV$$

$$1.73\ MeV$$

$$1.46\ MeV$$

$$Depends\ on\ the\ binding\ energy\ of\ proton.$$

Solution

Question 8

Kinetic energy of the bombarding neutron must be equal to the magnitude of the $$Q-value$$ of the reaction

Kinetic energy of the bombarding neutron must be less than the ,magnetic

Kinetic energy of the bombarding neutron must be greater than then magnetic

In this type of reaction, principle of conservation of energy is not valid

Question 9

$$0.24$$ MeV

$$1.44$$MeV

$$1.68$$MeV

$$3.12$$MeV

Solution

Question 10

1

2

3

5

Solution

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