Nuclei Test

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Nuclei Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

The Binding energy per nucleon of $$^7_3Li$$ and $$_2^4He$$ nucleon are $$5.60 MeV$$ and $$7.06 MeV$$, respectively. In the nuclear reaction $$_3^7Li+_1^1H\rightarrow _2^4He+^4_2He+Q$$, the value of energy $$Q$$ released is

A
$$19.6\ MeV$$

B
$$-2.4\ MeV$$

C
$$8.4\ MeV$$

D
$$17.3\ MeV$$

Solution

$$_4Li^7 + _1H^1 \rightarrow _2(_2He^4)$$

BE of products $$= (5.6 MeV) \times 7 + 0$$$$= 39.2 MeV$$
$$E = -39.2\ MeV$$

BE of reactant $$= (7.06) \times 4 \times 2$$$$= 56.48 MeV$$
$$E_f =- 56.48\ MeV$$

As nuclear energy decreases, so some energy will be released
$$Q_{release}=E_i-E_f=(-39.2)-(-56.48)=17.28\ MeV$$
Question 2
1 / -0

$$\displaystyle { S }^{ 32 }$$ absorbs energy and decays into which element after two $$\displaystyle \alpha $$-emissions?

A
Carbon

B
Aluminium

C
Oxygen

D
Magnesium

Solution

Two alpha emissions can be written as:
$$^{32}_{16}S\rightarrow ^{24}_{12}Mg+2^4_2He$$

Thus, sulphur decays into magnesium.
Question 3
1 / -0

Find $$BE$$ per nucleon of $$^{56}Fe$$ where $$m(^{56}Fe) = 55.936u\ m_{n} = 1.00727u, m_{p} = 1.007274\ u$$.

A
$$477.45\ MeV$$

B
$$8.52\ MeV$$

C
$$577\ MeV$$

D
$$10.52\ MeV$$

Solution

We know that ,
Binding Energy , $$BE=\Delta mc^2$$ where $$\Delta m $$ is the mass defect

Here , $$\Delta m =\text{Mass of nucleon- Mass of nucleus}= 26\ m_p + 30\ m_n - m^{56}Fe$$

Hence
$$BE = [26\ m_p + 30\ m_n - m^{56}Fe]c^{2}$$
Where , $$m_{n} = 1.00866\ u$$
$$m_{p} = 1.00727\ u$$

$$BE = [26\times 1.00727 + 30 \times 1.00866 - 55.936]\times 931$$
$$= 0.51282\times 931$$
$$= 477.435\ MeV$$
Binding energy per nucleon $$= \dfrac {477.435}{56}MeV = 8.52\ MeV$$.
Question 4
1 / -0

The mass of $$_{7}N^{15}$$ is $$15.0011\ amu$$, mass of $$_{8}O^{16}$$ is $$15.99492\ amu$$ and $$m_{P} = 1.00783\ amu$$. Determine binding energy of last proton of $$_{8}O^{16}$$.

A
$$2.13\ MeV$$

B
$$0.13\ MeV$$

C
$$10\ MeV$$

D
$$12.13\ MeV$$

Solution

$$M(_{8}O_{16}) = M(_{7}N^{15}) + 1m_{P}$$
binding energy of last proton
$$= M(N^{15}) + m_{P} - M(_{1}O^{16})$$
$$= 15.00011 + 1.00783 - 15.99492$$
$$= 0.01302\ amu = 12.13\ MeV$$.
Question 5
1 / -0

When two deuterium nuclei fuse together to form a tritium nucleus, we get a

A
neutron

B
deuteron

C
alpha particle

D
proton

Solution

Here the nuclear reaction : $$2 _1^2H(deutron) \rightarrow _1^3H (tritium)$$
For balancing the reaction : $$2 _1^2H \rightarrow _1^3H+ _1^1H(proton)$$
Thus, we will get a proton in this reaction.
Question 6
1 / -0

If $$M$$ is atomic weight , $$A$$ is mass number then $$\dfrac{M-A}{A}$$ represents :

A
Mass defect

B
Packing fraction

C
Binding energy

D
Chain reaction

Solution

Packing fraction is the mass defect per nucleon i.e. elementary particle in the nucleus. The difference between atomic weight and mass number i.e. mass of elementary particle in the nucleus is known as mass defect. Hence,
$$p = \dfrac{\Delta m}{A}$$
$$p = \dfrac{M - A}{A}$$
Question 7
1 / -0

Which of the following is wrong statement about binding energy ?

A
It is the sum of the rest mass energies of nucleons minus the rest mass energy of the nucleus

B
It is the energy released when the nucleons combine to form a nucleus

C
It is the energy required to break a given nucleus into its constituent nucleons

D
It is the sum of the kinetic energies of all the nucleons in the nucleus

Solution

The first three options are correct from the definition of Binding energy.
B.E. has nothing to do with K.E. of the nucleons in nucleus.
Question 8
1 / -0

When $$_{15}P^{30}$$ decays to become $$_{14}Si^{30}$$, which particle is released ?

A
electron

B
$$\alpha$$-particle

C
neutron

D
positron

Solution

The nuclear reaction : $$_{15}P^{30}\rightarrow$$ $${14}Si^{30} + $$ $$_{+1}e^0$$
Thus a positron is emitted during the decay of $$_{15}P^{30}$$ into $$_{14}Si^{30}$$.
Question 9
1 / -0

The difference between the mass of a nucleus and the combined mass of its nucleons is :

A
zero

B
positive

C
negative

D
zero, positive or negative

Solution

We know that mass defect $$=$$ combined mass of nucleons $$-$$ mass of the nucleus.
Since mass defect is always positive quantity so the difference of nucleus and the combined mass of its nucleons will be negative. The combined mass is greater than the mass of nucleus.
Question 10
1 / -0

What parameter is used to measure the stability of a nucleus?

A
Average binding energy

B
No. of protons

C
No. of neutrons

D
No. of electrons

Solution

Stability of nucleus is based on average binding energy i.e. binding energy per nucleon. This much energy will be needed for nucleon to break free.

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Re-Attempt Weekly Quiz Competition

Nuclei Test - 12

Result

Nuclei Test - 12

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SHARING IS CARING

Question 1

$$19.6\ MeV$$

$$-2.4\ MeV$$

$$8.4\ MeV$$

$$17.3\ MeV$$

Solution

Question 2

Carbon

Aluminium

Oxygen

Magnesium

Solution

Question 3

$$477.45\ MeV$$

$$8.52\ MeV$$

$$577\ MeV$$

$$10.52\ MeV$$

Solution

Question 4

$$2.13\ MeV$$

$$0.13\ MeV$$

$$10\ MeV$$

$$12.13\ MeV$$

Solution

Question 5

neutron

deuteron

alpha particle

proton

Solution

Question 6

Mass defect

Packing fraction

Binding energy

Chain reaction

Solution

Question 7

It is the sum of the rest mass energies of nucleons minus the rest mass energy of the nucleus

It is the energy released when the nucleons combine to form a nucleus

It is the energy required to break a given nucleus into its constituent nucleons

It is the sum of the kinetic energies of all the nucleons in the nucleus

Solution

Question 8

electron

$$\alpha$$-particle

neutron

positron

Solution

Question 9

zero

positive

negative

zero, positive or negative

Solution

Question 10

Average binding energy

No. of protons

No. of neutrons

No. of electrons

Solution

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