Nuclei Test

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Nuclei Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

The binding energy per nucleon of $$_5B^{10}$$ is $$8.0$$MeV and that of $$_5B^{11}$$ is $$7.5$$MeV. The Energy required to remove a neutron from $$_5B^{11}$$ is (mass of electron and proton are $$9.11\times 10^{-31}$$kg and $$1.67\times 10^{-27}$$kg).

A
$$2.5$$MeV

B
$$8.0$$MeV

C
$$0.5$$MeV

D
$$7.5$$MeV

Solution

The nuclear reaction is $$_5B^{11}\rightarrow_5B^{10} + _0n^1 $$
Binding energy per nucleon of $$_5B^{11}$$ $$E'_1= 7.5 MeV$$
Its binding energy $$E_1 = 7.5\times 11 = 82.5 MeV$$
Binding energy per nucleon of $$_5B^{10}$$ $$E'_2= 8.0 MeV$$
Its binding energy $$E_2 = 8.0\times 10 = 80 MeV$$
Energy required to remove a neutron $$\Delta E = E_1 - E_2$$
$$\therefore$$ $$\Delta E = 82.5 -80 = 2.5 MeV$$
Question 2
1 / -0

Two radioactive sources $$X$$ and $$Y$$ of half lives $$1h$$ and $$2h$$ respectively initially contain the same number of radioactive atoms. At the end of $$2h$$, their rates of disintegration are in the ratio of :

A
$$4:3$$

B
$$3:4$$

C
$$1:2$$

D
$$2:1$$

Solution

Rate of disintegration $$\alpha$$ number of atoms left
For $$X \dfrac{N}{N_0} = \left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}$$
For $$Y \dfrac{N}{N_0} = \left(\dfrac{1}{2}\right)^1 = \dfrac{1}{2}$$
$$\therefore \dfrac{R_X}{R_Y} = \dfrac{\dfrac{N_0}{4}}{\dfrac{N_0}{2}}=\dfrac{1}{2}$$
Question 3
1 / -0

The mass defect of $$_{ 2 }^{ 4 }{ He }$$ is $$0.03 u$$. The binding energy per nucleon of helium (in $$MeV$$) is

A
$$69.825$$

B
$$6.9825$$

C
$$2.793$$

D
$$27.93$$

Solution

Mass defect of $$^4_2He$$, $$\Delta M = 0.03 u$$
Binding energy of helium $$E = \Delta M \times 931$$ $$MeV$$
$$ E =0.03\times 931.5 =27.93 MeV$$
Number of nucleons in helium $$n =4$$
Binding energy per nucleon $$\dfrac{E}{n} = \dfrac{27.93}{4} =6.9825$$
Question 4
1 / -0

Atomic weight of boron is $$10.81$$ and it has two isotopes $$_{5}B^{10}$$ and $$_{5}B^{11}$$. Then ratio of $$_{5}B^{10} : \,_{5}B^{11}$$ in nature would be

A
$$19 : 81$$

B
$$10 : 11$$

C
$$15 : 16$$

D
$$81 : 19$$

Solution

Let the percentage of $$B^{10}$$ atoms be $$x$$, then average atomic weight
$$= \dfrac {10x + 11(100 - x)}{100} = 10.81$$
$$\Rightarrow x = 19$$
$$\therefore \dfrac {N_{B^{10}}}{N_{B^{11}}} = \dfrac {19}{81}$$.
Question 5
1 / -0

Nuclear reactor in which $$U - 235$$ is used as fuel. Uses $$2\ kg$$ of $$U-235$$ in $$30$$ days. Then, power output of the reactor will be (given energy released per fission $$= 185\ MeV$$).

A
$$43.5\ MW$$

B
$$58.5\ MW$$

C
$$69..6\ MW$$

D
$$73.1\ MW$$

Solution

Number of atoms in $$2\ kg$$ of uranium
$$= \dfrac {6.02\times 10^{23}}{235} \times 2000$$
$$= 5.12\times 10^{24}$$
Therefore, energy obtained from these atoms
$$= 5.12\times 10^{24} \times 185\ MeV$$
$$= 5.12\times 10^{24}\times 185\times 10^{6}eV$$
Energy obtained per second
$$= \dfrac {5.12\times 10^{24}\times 185\times 10^{6} \times 16\times 10^{-19}}{30\times 24\times 60\times 60}$$
Solving the above expression
$$= 58.47\times 10^{6}W$$
$$\equiv 58.5\ MW$$.
Question 6
1 / -0

The binding energy per nucleon of $$C^{12}$$ is $$E_{1}$$ and that of $$C^{13}$$ is $$E_{2}$$. The energy required to remove one neutron from $$C^{13}$$ is

A
$$12E_{1} + 12E_{2}$$

B
$$13E_{2} - 12E_{1}$$

C
$$12E_{2} - 13E_{1}$$

D
$$13E_{2} + 12E_{1}$$

Solution

Binding energy of $$C^{12}$$ is
$$BE_{1} = 12\times E_{1}$$
Binding energy of $$C^{13}$$ is
$$BE_{2} = 13\times E_{2}$$
Thus, energy required to remove one neutron from $$C^{13}$$ is
$$\triangle E = BE_{2} - BE_{1} = 13E_{2} - 12E_{1}$$.
Question 7
1 / -0

A nucleus X initially at rest, undergoes alpha decay according to the equation.
$$_{92}X^A\rightarrow _{Z}Y^{228}+\alpha$$.
Then, the value of A and Z are.

A
$$94,230$$

B
$$232,90$$

C
$$190,32$$

D
$$230,94$$

Solution

The decay equation is
$$_{92}X^A\rightarrow _ZY^{228}+\alpha$$
$$\alpha$$-particle is nucleus of $$_2He^4$$
$$_{92}X^A\rightarrow _ZY^{228}+_2He^4$$
$$A=228+4=232$$
$$Z=92-2=90$$.
Question 8
1 / -0

Li nucleus has three protons and four neutrons. Mass of lithium nucleus is 7.03.6005 amu. Mass of proton is 1.007277 amu and mass of neutron is 1.008665 amu. Mass defect for lithium nucleus in amu is

A
0.04048

B
0.04050

C
0.04052

D
0.04055

Solution

Mass defect = mass of nucleons - mass of nucleus
$$ = (3 \times 1.007277 + 4 t008665)- 7.016005$$
$$ = 0.040486\, amu $$
$$\approx 0.04050 $$
Question 9
1 / -0

Nucleus of mass number $$A$$, originally at rest, emits the $$\alpha$$ particle with speed $$v$$. The daughter nucleus recoils with a speed of :

A
$$2v(A+4)$$

B
$$4v(A+4)$$

C
$$4v/(A-4)$$

D
$$2v/(A-4)$$

Solution

According to conservation of momentum
$$4v=(a-4)v'\Rightarrow v'=\cfrac { 4v }{ A-4 } $$
Question 10
1 / -0

The binding energy per nucleon of deuteron $$(_{1}H^{2})$$ and helium $$(_{2}He^{4})$$ are $$1.1\ MeV$$ and $$7.0\ MeV$$, respectively. The energy released when two deuterons fuse to form a helium nucleus is

A
$$36.2\ MeV$$

B
$$23.6\ MeV$$

C
$$47.2\ MeV$$

D
$$11.8\ MeV$$

E
$$9.31\ MeV$$

Solution

The nuclear reaction is,
$$_{1}H^{2} + _{1}H^{2} \rightarrow _{2}He^{4} + Q$$
Total binding energy of helium nucleus
$$= 4\times 7 = 28\ MeV$$
Total binding energy of each deuteron
$$= 2\times 1.1 = 2.2\ MeV$$
Hence, energy released when two deuteron fuse to form helium,
$$= 28 - 2 \times 2.2 = 28 - 44 = 23.6\ MeV$$.

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Nuclei Test - 43

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Nuclei Test - 43

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Question 1

$$2.5$$MeV

$$8.0$$MeV

$$0.5$$MeV

$$7.5$$MeV

Solution

Question 2

$$4:3$$

$$3:4$$

$$1:2$$

$$2:1$$

Solution

Question 3

$$69.825$$

$$6.9825$$

$$2.793$$

$$27.93$$

Solution

Question 4

$$19 : 81$$

$$10 : 11$$

$$15 : 16$$

$$81 : 19$$

Solution

Question 5

$$43.5\ MW$$

$$58.5\ MW$$

$$69..6\ MW$$

$$73.1\ MW$$

Solution

Question 6

$$12E_{1} + 12E_{2}$$

$$13E_{2} - 12E_{1}$$

$$12E_{2} - 13E_{1}$$

$$13E_{2} + 12E_{1}$$

Solution

Question 7

$$94,230$$

$$232,90$$

$$190,32$$

$$230,94$$

Solution

Question 8

0.04048

0.04050

0.04052

0.04055

Solution

Question 9

$$2v(A+4)$$

$$4v(A+4)$$

$$4v/(A-4)$$

$$2v/(A-4)$$

Solution

Question 10

$$36.2\ MeV$$

$$23.6\ MeV$$

$$47.2\ MeV$$

$$11.8\ MeV$$

$$9.31\ MeV$$

Solution

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