Nuclei Test

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Nuclei Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

A nuclear transformation is denoted by X(n,$$\alpha $$)$$\rightarrow _{3} Li^{7}$$. The nucleus of element X is

A
$$_{6}C^{12}$$

B
$$_{5}B^{10}$$

C
$$_{5}B^{9}$$

D
$$_{4}Be^{11}$$

Solution

The notation means that on being bombarded with neutrons the element breaks into Lithium and an $$\alpha-particle.$$
$$^{Z}_{A}X + ^{1}_{0}n \Rightarrow ^{Z-3}_{A-2}Y + ^{4}_{2}He$$

Now,
$$^{Z-3}_{A-2}Y = ^{7}_{3}Li$$

$$\therefore Z = 10 , A= 5 $$
So, The element is Boron and X is $$_{5}B^{10} $$
Question 2
1 / -0

$$20$$% of a radioactive element disintegrates in $$1hr$$.The percentage of the radioactive element disintegrated in $$2hrs$$ will be

A
$$36$$%

B
$$64$$%

C
$$60$$%

D
$$40$$%

Solution

Let initially the number of radioactive particles be $$100$$.
Using the formula:
$$N = N_0 e^{-\lambda t}$$ where $$\lambda = \dfrac{2.303}{t} log(\dfrac{N_0}{N})$$

$$\lambda = \dfrac{2.303}{1} log(\dfrac{100}{80}) \Rightarrow \lambda = 0.223 hrs^{-1}$$

Radioactive material left in $$2\ hrs$$ $$ = N_0 e^{-\lambda t}$$
$$ = 100 e^{-0.223\times2}$$
$$ =63.99 \approx 64%$$

Radioactive material decayed in 2 hours $$ = (100-64)%$$
$$ = 36$$ %
Question 3
1 / -0

Assertion (A) : Isotopes of an element can be separated by using a mass spectrometer.
Reason (R) : Separation of isotopes is possible due to difference in electron number of isotopes.

A
Both A and R are true and R is correct explanation of A

B
Both A and R are true and R is not correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

This is fact that isotopes can be separated using mass spectrometer and 2nd statement is false because in isotopes the number of neutrons differs not electrons.
Question 4
1 / -0

Consider the following statements (A) and (B) and identify the correct answer given below :
Statement A: Positive values of packing fraction implies a large value of binding energy
Statement B: The difference between the mass of the nucleus and the mass number of the nucleus is called the packing fraction

A
A and B are correct

B
A and B are false

C
A is true, B is false

D
A is false, B is true

Solution

Packing fraction $$= \dfrac{Actual \ isotopic \ mass - mass \ number}{ Mass\ number}$$

A negative packing fraction indicates stability of the nucleus thus a large Binding energy and vice versa.
So, we easily see that statement A and b are false.
Question 5
1 / -0

True masses of neutron,proton and deutron in $$a.m.u$$ are $$1.00893, 1.00813$$ and $$2.01473$$ respectively. The packing fraction of the deutron in $$a.m.u$$ is

A
$$11.56 \times 10^{-4}$$

B
$$23.5 \times 10^{-4}$$

C
$$33.5 \times 10^{-4}$$

D
$$47.15 \times 10^{-4}$$

Solution

Deutron $$=\ ^{2}_{1}H$$ i.e., $$1$$ proton and $$1$$ neutron.

Packing fraction $$= \dfrac{True \ mass - Mass \ number}{Mass\ number}$$

$$ = \dfrac{(1.00893 + 1.00813) - 2.01473}{2.01473} $$

$$ = 11.56 \times10^{-4} $$
Question 6
1 / -0

$$4.6\times 10^{22}$$ atoms of an element weight $$13.8$$ g. What is the atomic mass of the element?

A
290 u

B
180.6 u

C
34.4 u

D
104 u

Solution

$$1$$ mole of any substance contains $$6.02 \times 10^{23}$$ atoms.

Thus, $$4.6 \times 10^{22}$$ atoms corresponds to $$\dfrac{4.6 \times 10^{22}}{6.022 \times 10 ^{23}}=0.0764$$ moles.

$$0.0764$$ moles weighs $$13.8$$ g.

Thus, $$1$$ mole will weigh $$\dfrac{13.8}{0.0764}= 180.6 \ g$$.

Hence, the atomic mass of the element will be 180.6 u.
Question 7
1 / -0

When the mass of an electron becomes equal to thrice its rest mass, its speed is

A
$$\dfrac{2\sqrt{2}}{3}c$$

B
$$\dfrac{2}{3}c$$

C
$$\dfrac{1}{3}c$$

D
$$\dfrac{1}{4}c$$

Solution

The mass of a moving body $$m$$ is given by special theory of relativity is given by:
$$m=\dfrac{m_0}{\sqrt{1-\dfrac{v^2}{c^2}}}$$,
where, $$m_0$$ is the rest mass of the body.
According to the question, the mass of an electron becomes equal to thrice its rest mass,
So, $$m=3m_0$$.
Using this, the speed of the moving body will be: $$v=\dfrac{2\sqrt {2}}{3} c$$
Question 8
1 / -0

Assertion (A) : If a heavy nucleus is split into two medium sized parts,each of the new nuclei will have more binding energy per nucleon than the original nucleus.
Reason (R): Joining two light nuclei together to give a single nucleus of medium size means more binding energy per nucleon in the new nucleus.

A
Both A and R are true and R is correct explanation of A

B
Both A and R are true and R is not correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

If a heavy nucleus splits, it forms smaller medium size particles, which are more stable as compared to parent nuclei thus having higher Binding energy. And this is the principle for nuclear fission and in case of nuclear fusion the two light nuclei join to form a more stable medium sized nuclei to get the optimum n/p ratio as close and near to Fe which has most stable nuclei in terms of n/p ratio.
Question 9
1 / -0

An electron moves with a speed of $$\dfrac{\sqrt{3}}{2}c$$. Then its mass becomes_____ times its rest mass. (Given velocity of light $$=c$$)

A
$$2$$

B
$$3$$

C
$$3/2$$

D
$$4$$

Solution

Mass of on electron moving with speed $$V$$ is $$\dfrac{m_o}{\sqrt{1-\dfrac{v^{2}}{c^{2}}}}$$
where, $$m_o$$ is rest mass
$$V$$ is speed of electron
$$C$$ is speed of light

So, $$m$$ $$=\dfrac{m_o}{\sqrt{1-\dfrac{v^{2}}{c^{2}}}}$$

$$=\dfrac{m_o}{\sqrt{1-\dfrac{\left ( \frac{\sqrt{3C}}{2} \right )^{2}}{C^{2}}}}$$

$$=\dfrac{m_o}{\sqrt{1-\dfrac{3}{4}}}$$

$$=\dfrac{m_o}{\sqrt{\frac{1}{4}}}$$

$$=\dfrac{m_o}{\dfrac{1}{2}}$$

$$=2m_o$$

So, mass becomes 2 times of its rest mass.
Question 10
1 / -0

Assertion (A): Nuclear fusion reactions are considered as thermo-nuclear reactions
Reason (R): The source of stellar energy is nuclear fusion

A
Both A & R are true and R is the correct explanation of A

B
Both A & R are true and R is not correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

Nuclear fusion reaction are consider as thermo-nuclear reaction because for the nuclear fusion reaction to start we have to provide lot of temperature and heat energy to overcome the potential energy before fusion and the source of energy of stars is nothing but nuclear energy.

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Nuclei Test - 26

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Nuclei Test - 26

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SHARING IS CARING

Question 1

$$_{6}C^{12}$$

$$_{5}B^{10}$$

$$_{5}B^{9}$$

$$_{4}Be^{11}$$

Solution

Question 2

$$36$$%

$$64$$%

$$60$$%

$$40$$%

Solution

Question 3

Both A and R are true and R is correct explanation of A

Both A and R are true and R is not correct explanation of A

A is true but R is false

A is false but R is true

Solution

Question 4

A and B are correct

A and B are false

A is true, B is false

A is false, B is true

Solution

Question 5

$$11.56 \times 10^{-4}$$

$$23.5 \times 10^{-4}$$

$$33.5 \times 10^{-4}$$

$$47.15 \times 10^{-4}$$

Solution

Question 6

290 u

180.6 u

34.4 u

104 u

Solution

Question 7

$$\dfrac{2\sqrt{2}}{3}c$$

$$\dfrac{2}{3}c$$

$$\dfrac{1}{3}c$$

$$\dfrac{1}{4}c$$

Solution

Question 8

Both A and R are true and R is correct explanation of A

Both A and R are true and R is not correct explanation of A

A is true but R is false

A is false but R is true

Solution

Question 9

$$2$$

$$3$$

$$3/2$$

$$4$$

Solution

Question 10

Both A & R are true and R is the correct explanation of A

Both A & R are true and R is not correct explanation of A

A is true but R is false

A is false but R is true

Solution

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