M.C.Q (1 Marks) - Physics STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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MCQ 11 Mark

Ten grams of $^{57}Co$ kept in an open container beta$-$decays with a half$-$life of $270$ days. The weight of the material inside the container after $540$ days will be very nearly$:$

✓
$10g$
B
$5g$
C
$2.5g$
D
$1.25g$

Answer

Correct option: A.

$10g$

$^{57}Co$ is undergoing beta decay, i.e. electron is being produced. But an electron has very less mass $(9.11 \times 10^{-31}kg)$ as compared to the $Co$ atom.
Therefore, after $570$ days, even though the atoms undergo large beta decay, the weight of the material in the container will be nearly $10g.$

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Question 21 Mark

The mass of a neutral carbon atom in ground state is:

Exact 12u
Less than 12u
More than 12u
Depends on the form of carbon such as graphite or charcoal.

Answer

Exact 12u

Explanation:
In nuclear physics, a unit used for measurement of mass is unified atomic mass unit, which is denoted by u.
It is defined such that
$1\text{u}=\frac{1}{12}\times$ (Mass of neutral carbon atom in its ground state)
Mass of neutral carbon atom in its ground state = 12 × 1u = 12u
Thus, the mass of neutral carbon atom in its ground state is exactly 12u.

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Question 31 Mark

Magnetic field does not cause deflection in:

$\alpha-\text{rays}$
$\beta^+-\text{rays}$
$\beta^--\text{rays}$
$\gamma-\text{rays}$

Answer

$\gamma-\text{rays}$

Explanation:
Magnetic force acts on a charged particle, due to which it deflects from its path. The magnitude of this force is measured as $\Big|\overrightarrow{\text{F}}\Big|=\Big|\text{q}\Big(\overrightarrow{\text{v}}\times\overrightarrow{\text{B}}\Big)\Big|.$
Here, q is the charge on the particle that is moving with speed v in a uniform magnetic field B.
Since alpha, beta-plus and beta-minus are charged particles, they suffer deflection due to the field applied. On the other hand, gamma rays are photons and due to zero charge, they do not suffer any deflection.

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MCQ 41 Mark

As compared to $^{12}C$ atom $,^{14}C$ atom has:

A
Two extra protons and two extra electrons.
B
Two extra protons but no extra electron.
✓
Two extra neutrons and no extra electron.
D
Two extra neutrons and two extra electrons.

Answer

Correct option: C.

Two extra neutrons and no extra electron.

$^{12}C$ and $^{14}C$ are the two isotopes of carbon atom that have same atomic number, but different mass numbers. This means that they have same number of protons and electrons, but different number of neutrons.
Therefore $^{12}C$ has $6$ protons$, 6$ electrons and $6$ neutrons, whereas $^{14}C$ has $6$ electrons$, 6$ protons and $8$ neutrons.

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MCQ 51 Mark

The mass number of a nucleus is:

A
Always less than its atomic number.
B
Always more than its atomic number.
C
Equal to its atomic number.
✓
Sometimes more than and sometimes equal to its atomic number.

Answer

Correct option: D.

Sometimes more than and sometimes equal to its atomic number.

Mass number of a nucleus is defined as the sum of the number of neutron and protons present in the nucleus, i.e. the number of nucleons in the nucleus, whereas atomic number is equal to the number of protons present. Therefore, the atomic number is smaller than the mass number. But in the nucleus $($like that of hydrogen $^1H_{1}),$ only protons are present. Due to this, the mass number is equal to the atomic number.

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MCQ 61 Mark

A freshly prepared radioactive source of half$-$life $2h$ emits radiation of intensity which is $64$ times the permissible safe level. The minimum time after which it would be possible to work safely with this source is:

A
$6h$
✓
$12h$
C
$24h$
D
$128h$

Answer

Correct option: B.

$12h$

A freshly prepared radioactive source emits radiation of intensity that is $64$ times the permissible level.
This means that it is possible to work safely till $6$ half $-$ lives $($as $2^6 = 64)$ of the radioactive source.
Since the half $-$ life of the source is $2h,$ the minimum time after which it would be possible to work safely with this source is $12h$.

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Question 71 Mark

In which of the following decays the element does not change?

$\alpha-\text{decay}$
$\beta^+-\text{decay}$
$\beta^--\text{decay}$
$\gamma-\text{decay}$

Answer

$\gamma-\text{decay}$

Explanation:
In alpha particle decay, the unstable nucleus emits an alpha particle reducing its proton number Z by 4 and neutron number N by 2 such that the element gets changed.
$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}-4}_{\text{Z}-2}\text{Y}+\text{ }^4_2\text{He}$
During $\beta^--\text{decay},$ a neutron is converted to a proton, an electron and an antineutrino, i.e. an active nucleus gets converted to one of its isobars and hence the element gets changed.
$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}}_{\text{Z}+1}\text{Y}+\text{e}+\bar{\text{v}}$
During $\beta^+-\text{decay},$ a proton in the nucleus is converted to a neutron, a positron and a neutrino in order to maintain the stability of the nucleus, i.e. an active nucleus gets converted to one of its isobars and hence the element gets changed.
$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}}_{\text{Z}-1}\text{Y}+\beta^++\text{v}$
When a nucleus is in higher excited state or has excess of energy, it comes to the ground state in order to become stable and release energy in the form of electromagnetic radiation called gamma ray. Hence, the element in gamma decay doesn't change.

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MCQ 81 Mark

Free $^{238}U$ nuclei kept in a train emit alpha particles. When the train is stationary and a uranium nucleus decays, a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes $x$ in time $t$ after the decay. If a decay takes place when the train is moving at a uniform speed $v,$ the distance between the alpha particle and the recoiling nucleus at a time $t$ after the decay, as measured by the passenger will be:

A
$x + vt$
B
$x - vt$
✓
$x$
D
depends on the direction of the train.

Answer

Correct option: C.

$x$

When the train is stationary, the separation between the alpha particle and recoiling uranium nucleus is $x$ in time $t$ after the decay.
Even if the decay is taking place in a moving train and the separation is measured by the passenger sitting in it, the separation between the alpha particle and nucleus will be $x$.
This is because the observer is also moving with the same speed with which the alpha particle and recoiling nucleus are moving, i.e. they all are in the same frame that is moving at a uniform speed.

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Question 91 Mark

The heavier nuclei tend to have larger $\frac{\text{N}}{\text{Z}}$ ratio because:

A neutron is heavier than a proton.
A neutron is an unstable particle.
A neutron does not exert electric repulsion.
Coulomb forces have longer range compared to the nuclear forces.

Answer

A neutron does not exert electric repulsion.
Coulomb forces have longer range compared to the nuclear forces.

Explanation:
This is because in heavy nuclei, the $\frac{\text{N}}{\text{Z}}$ ratio becomes larger in order to maintain their stability and reduce instability caused due to the repulsion among the protons. The neutrons exert only attractive short-range nuclear forces on each other as well as on the neighbouring protons, whereas the protons exert attractive short-range nuclear forces on each other as well as the electrostatic repulsive force. Thus, the nuclei with high mass number, in order to be stable, have large neutron to proton ratio $\frac{\text{N}}{\text{Z}}.$

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MCQ 101 Mark

As the mass number A increases, which of the following quantities related to a nucleus do not change?

A
Mass.
B
Volume.
✓
Density.
D
Binding energy.

Answer

Correct option: C.

Density.

Radius of a nucleus with mass number $A$ is given as
$\text{R}=\text{R}_{\text{0}}\text{A}^{\frac{1}{3}}$
Here, $\text{R}_0=1.2\text{fm}$
$\therefore$ Volume of the nucleus $=\frac{4\pi\text{R}^3}{3}=\frac{4\pi\text{R}^3\text{A}}{3}$
This depends on $A.$ With an increase in $A, V$ increases proportionally.
Mass of the nucleus $\simeq\text{Am}_{\text{N}}$
Here, $m_{N }$ is the mass of a nucleon.
Therefore, mass of the nucleus also increases with the increasing mass number. Binding energy also depends on mass number $($number of nucleons$)$ as it is the difference between the total mass of the constituent nucleons and the nucleus. Therefore, it also varies with the changing mass number.
On the other hand,
$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$
$=\frac{\text{Am}_{\text{N}}}{\frac{4\pi\text{R}3}{3}}=\frac{\text{Am}_{\text{N}}}{\frac{4\pi\text{R}_0^3\text{A}}{3}}=\frac{\text{m}_{\text{N}}}{\frac{4\pi\text{R}_0^3}{3}}=\frac{3\text{m}_{\text{N}}}{4\pi\text{R}_{0}^3}$
This is independent of $A$ and hence does not change as mass number increases.

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Question 111 Mark

In which of the following decays the atomic number decreases?

$\alpha-\text{decay}$
$\beta^+-\text{decay}$
$\beta^--\text{decay}$
$\gamma-\text{decay}$

Answer

$\alpha-\text{decay}$
$\beta^+-\text{decay}$

Explanation:
In alpha particle decay, the unstable nucleus emits an alpha particle reducing its proton number (atomic number) Z as well as neutron number N by 2.
$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}-4}_{\text{Z}-2}\text{Y}+\text{ }^4_2\text{He}$
During $\beta^--\text{decay},$ a neutron is converted to a proton, an electron and an antineutrino. Thus, there is an increase in the atomic number.
$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}}_{\text{Z}+1}\text{Y}+\text{e}^-+\bar{\text{v}}$
During $\beta^+-\text{decay},$ a proton in the nucleus is converted to a neutron, a positron and a neutrino in order to maintain the stability of the nucleus. Thus, there is a decrease in the atomic number.
$\text{ }^{\text{A}}_{\text{Z}}\text{X}\rightarrow\text{ }^{\text{A}}_{\text{Z}-1}\text{Y}+\beta^++\text{v}$
When a nucleus is in higher excited state or has excess of energy, it comes to the lower state in order to become stable and release energy in the form of electromagnetic radiation called gamma ray. The element in the gamma decay doesn't change.
Therefore, alpha and beta plus decay suffer decrease in atomic number.

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Question 121 Mark

As the mass number A increases, the binding energy per nucleon in a nucleus:

Increases.
Decreases.
Remains the same.
Varies in a way that depends on the actual value of A.

Answer

Varies in a way that depends on the actual value of A.

Explanation:
Binding energy per nucleon in a nucleus first increases with increasing mass number (A) and reaches a maximum of 8.7MeV for A (50 - 80). Then, again it slowly starts decreasing with the increase in A and drops to the value of 7.5MeV.

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Question 131 Mark

Consider a sample of a pure beta-active material:

All the beta particles emitted have the same energy.
The beta particles originally exist inside the nucleus and are ejected at the time of beta decay.
The antineutrino emitted in a beta decay has zero mass and hence zero momentum.
The active nucleus changes to one of its isobars after the beta decay.

Answer

The active nucleus changes to one of its isobars after the beta decay.

Explanation:
In a beta decay, either a neutron is converted to a proton or a proton is converted to a neutron such that the mass number does not change. Also, the number of the nucleons present in the nucleus remains the same. Thus, the active nucleus gets converted to one of its isobars after beta decay.

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Question 141 Mark

Which of the following are electromagnetic waves?

$\alpha-\text{decay}$
$\beta^+-\text{decay}$
$\beta^--\text{decay}$
$\gamma-\text{decay}$

Answer

$\gamma-\text{decay}$

Explanation:
Alpha rays, beta-plus and beta-minus rays carry charged particles that show particle behaviour. On the other hand, gamma rays carry photons that show particle as well as wave behaviour. Hence, only gamma rays are electromagnetic waves.

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Question 151 Mark

A free neutron decays to a proton but a free proton does not decay to a neutron. This is because:

Neutron is a composite particle made of a proton and an electron whereas proton is a fundamental particle.
Neutron is an uncharged particle whereas proton is a charged particle.
Neutron has large rest mass than the proton.
Weak forces can operate in a neutron but not in a proton.

Answer

Neutron has large rest mass than the proton.

Explanation:
A nucleus is made up of two fundamental particles-neutrons and protons. If a nucleus has more number of neutrons than what is needed to have stability, then neutrons decay into protons and if there's an excess of protons, then they decay to form neutrons. Since a neutron has larger rest mass than a proton, the Q-value of its decay reaction is positive and a free neutron decays to a proton, while an isolated proton cannot decay to a neutron as the Q-value of its decay reaction is negative. Hence, it is physically not possible.

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MCQ 161 Mark

An $\alpha-$particle is bombarded on $^{14}N.$ As a result $,a ^{17}O$ nucleus is formed and a particle is emitted. This particle is $a:$

A
Neutron.
✓
Proton.
C
Electron.
D
Positron.

Answer

Correct option: B.

Proton.

If an alpha particle is bombarded on a nitrogen $(N-14)$ nucleus, an oxygen $(O-17)$ nucleus and a proton are released.
According to the conservation of mass and charge,
$^4_2\text{He}+\text{ }^{14}_7\text{N}\rightarrow\text{ }^{17}_6\text{O}+\text{ }^1_1\text{p}$
So, the emitted particle is a proton.

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Question 171 Mark

Two lithium nuclei in a lithium vapour at room temperature do not combine to form a carbon nucleus because:

A lithium nucleus is more tightly bound than a carbon nucleus.
Carbon nucleus is an unstable particle.
It is not energetically favourable.
Coulomb repulsion does not allow the nuclei to come very close.

Answer

Coulomb repulsion does not allow the nuclei to come very close.

Explanation:
Lithium atom contains 3 protons and 3 neutrons in the nucleus and 3 valence electrons. When two lithium nuclei are brought together, they repel each other. The attractive nuclear forces being short-range are insignificant as compared to the electrostatic repulsion. Thus, the nuclei do not combine to form carbon atom because of coulomb repulsion.

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MCQ 181 Mark

The decay constant of a radioactive sample is $\lambda.$ The half-life and the average-life of the sample are respectively:

A
$\frac{1}{\lambda}$ and $\Big(\text{ln}\frac{2}{\lambda}\Big)$
✓
$\Big(\text{ln}\frac{2}{\lambda}\Big)$ and $\frac{1}{\lambda}$
C
$\lambda(\text{ln}2)$ and $\frac{1}{\lambda}$
D
$\frac{\lambda}{(\text{ln})2}$ and $\frac{1}{\lambda}$

Answer

Correct option: B.

$\Big(\text{ln}\frac{2}{\lambda}\Big)$ and $\frac{1}{\lambda}$

The half-life of a radioactive sample $\Big(\text{t}_{\frac{1}2{}}\Big)$ is defined as the time elapsed before half the active nuclei decays.
Let the initial number of the active nuclei present in the sample be $N_0.$
$\frac{\text{N}_{0}}{2}=\text{N}_{\text{0}}\text{e}^{-\lambda\text{t}_{\frac{1}2{}}}$
$\Rightarrow\text{t}_{\frac{1}{2}}=\frac{\text{In}2}{\lambda}$
Average life of the nuclei, $\text{t}_{\text{av}}=\frac{\text{S}}{\text{N}_{0}}=\frac{1}{\lambda}$
Here, $S$ is the sum of all the lives of all the $N$ nuclei that were active at $t = 0$ and $\lambda$ is the decay constant of the sample.

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MCQ 191 Mark

Let $F_{pp} , F_{pn}$ and $F_{nn}$ fill denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron respectively. When the separation is $1\ fm:$

A
$F_{pp} > F_{pn }= F_{nn}$
✓
$F_{pp} = F_{pn }= F_{nn}$
C
$F_{pp} > F_{pn }> F_{nn}$
D
$F_{pp} < F_{pn }= F_{nn}$

Answer

Correct option: B.

$F_{pp} = F_{pn }= F_{nn}$

Protons and neutrons are present inside the nucleus and they exert strong attractive nuclear forces on each other, which are equal in magnitude. Due to their positive charge, protons repel each other.
Hence the net attractive force between two protons gets reduced, but the nuclear force is stronger than the electrostatic force at a separation of $1\ fm.$

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Question 201 Mark

In a radioactive decay, neither the atomic number nor the mass number changes. Which of the following particles is emitted in the decay?

Proton.
Neutron.
Electron.
Photon.

Answer

Photon.

Explanation:
The atomic number and mass number of a nucleus is defined as the number of protons and the sum of the number of protons and neutrons present in the nucleus, respectively. Since in the decay, neither the atomic number nor the mass number change, it cannot be a beta-decay (release of electron, proton or neutron). Hence, the particle emitted can only be a photon.

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MCQ 211 Mark

Let $F_{pp,} F_{pn}$ and $F_{nn}$ denote the magnitudes of the net force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron respectively. Neglect gravitational force. When the separation is $1\ fm:$

A
$F_{pp} > F_{pn }= F_{nn}$
✓
$F_{pp} = F_{pn }= F_{nn}$
C
$F_{pp} > F_{pn }> F_{nn}$
D
$F_{pp} < F_{pn }= F_{nn}$

Answer

Correct option: B.

$F_{pp} = F_{pn }= F_{nn}$

Protons and neutrons are present inside the nucleus and they exert strong attractive nuclear force on each other.
These forces are equal in magnitude, irrespective of the charge present on the nucleons.
$\therefore F_{pp} = F_{pn }= F_{nn}$
Here$,_{ }F_{pp} = F_{pn }= F_{nn}$ denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron, respectively.

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Question 221 Mark

The mass number of a nucleus is equal to:

The number of neutrons in the nucleus.
The number of protons in the nucleus.
The number of nucleons in the nucleus.
None of them.

Answer

The number of nucleons in the nucleus.

Explanation:
Mass number of a nucleus is defined as the sum of the number of neutron and protons present in the nucleus, i.e. the number of nucleons in the nucleus.

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Question 231 Mark

The graph of ln$\Big(\frac{\text{R}}{\text{R}_{\text{0}}}\Big)$ versus In A (R = radius of a nucleus and A - its mass number) is:

A straight line.
A parabola.
An ellipse.
None of them.

Answer

A straight line.

Explanation:
The average nuclear radius (R) and the mass number of the element (A) has the following relation:
$\text{R}=\text{R}_{0}\text{A}^{\frac{1}{2}}$
$\frac{\text{R}}{\text{R}_{0}}=\text{A}^{\frac{1}{3}}$
In $\Big(\frac{\text{R}}{\text{R}_0}\Big)=\frac{1}{3}$ In A
Therefore, the graph of ln$\Big(\frac{\text{R}}{\text{R}_0}\Big)$ versus ln A is a straight line passing through the origin with slope $\frac{1}{3}.$

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MCQ 241 Mark

Two protons are kept at a separation of $10\ nm.$ Let $F_n$ and $F_e$ be the nuclear force and the electromagnetic force between them:

A
$F_e = F_n.$
✓
$F_e >> F_n.$
C
$F_e << F_n.$
D
$F_{e }$ and $F_n$ differ only slightly.

Answer

Correct option: B.

$F_e >> F_n.$

Two protons exert strong attractive nuclear force and repulsive electrostatic force on each other.
Nuclear forces are short range forces existing in the range of a few fms.
Therefore, at a separation of $10\ nm,$ the electromagnetic force is greater than the nuclear force, i.e. $F_e >> F_n.$

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Question 251 Mark

In one average-life:

Half the active nuclei decay.
Less than half the active nuclei decay.
More than half the active nuclei decay.
All the nuclei decay.

Answer

More than half the active nuclei decay.

Explanation:
The average life is the mean life time for a nuclei to decay.
It is given as
$\tau=\frac{1}{\lambda}=\frac{\text{T}_{\frac{1}{2}}}{0.693}$
Here, $\tau,\lambda$ and $\text{T}_{\frac{1}{2}}$ are the average life, decay constant and half life-time of the active nuclei, respectively. The value of the average lifetime comes to be more than the average lifetime. Therefore, in one average life, more than half the active nuclei decay.

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Question 261 Mark

During a nuclear fission reaction:

A heavy nucleus breaks into two fragments by itself a light nucleus bombarded by thermal neutrons breaks up.
A light nucleus bombarded by thermal neutrons breaks up.
A heavy nucleus bombarded by thermal neutrons breaks up.
Two light nuclei combine to give a heavier nucleus and possible other products.

Answer

A heavy nucleus bombarded by thermal neutrons breaks up.

Explanation:
In a nuclear reactor, a large fissile atomic nucleus like uranium-235 absorbs a thermal neutron and undergoes a nuclear fission reaction. The heavy nucleus splits into two or more lighter nuclei releasing gamma radiation and free neutrons.

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Question 271 Mark

Which of the following is a wrong description of binding energy of a nucleus?

It is the energy required to break a nucleus into its constituent nucleons.
It is the energy made available when free nucleons combine to form a nucleus.
It is the sum of the rest mass energies of its nucleons minus the rest mass energy of the nucleus.
It is the sum of the kinetic energy of all the nucleons in the nucleus.

Answer

It is the sum of the kinetic energy of all the nucleons in the nucleus.

Explanation:
Binding energy of a nucleus is defined as the energy required to break the nucleus into its constituents. It is also measured as the Q-value of the breaking of nucleus, i.e. the difference between the rest energies of reactants (nucleus) and the products (nucleons) or the difference between the kinetic energies of the products and the reactants.

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Question 281 Mark

During a negative beta decay:

An atomic electron is ejected.
An electron which is already present within the nucleus is ejected.
A neutron in the nucleus decays emitting an electron.
A proton in the nucleus decays emitting an electron.

Answer

A neutron in the nucleus decays emitting an electron.

Explanation:
Negative beta decay is given as
$\text{n}\rightarrow\text{p + e}^-+\bar{\text{v}}$
Neutron decays to produce proton, electron and anti-neutrino.

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Question 291 Mark

For nuclei with A > 100:

The binding energy of the nucleus decreases on an average as A increases.
The binding energy per nucleon decreases on an average as A increases.
If the nucleus breaks into two roughly equal parts, energy is released.
If two nuclei fuse to form a bigger nucleus, energy is released.

Answer

The binding energy per nucleon decreases on an average as A increases.
If the nucleus breaks into two roughly equal parts, energy is released.

Explanation:
Binding energy per nucleon varies in a way that it depends on the actual value of mass number (A). As the mass number (A) increases, the binding energy also increases and reaches its maximum value of 8.7MeV for A(50-80) and for A > 100. The binding energy per nucleon decreases as A increases and the nucleus breaks into two or more atoms of roughly equal parts so as to attain stability and binding energy of mass number between 50-80.

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M.C.Q (1 Marks) - Physics STD 12 Science Questions - Vidyadip