MCQ Questions for Class 12 Medical Physics Nuclei Quiz 4

The binding energy of an

$\alpha$ -particle is
(Given that mass of proton

$=$

$1.0073$ amu, mass of neutron

$=$

$1.0087$ amu, and mass of

$\alpha$ -particle

$=$

$4.0015$ amu)

0%
$28.4 MeV$
0%
$35.6 MeV$
0%
$40.3 MeV$
0%
$44.7 MeV$

Explanation

$\Delta m = (2\times 1.0073 + 2\times 1.0087 - 4.0015)$

$= 0.00305 \mu$

$E = 0.0305\times 931.478 MeV$

$= 28.41\ MeV$

The binding energy per nucleon of Uranium in

$MeV$ is
(Atomic mass of uranium

$m_{a}=\ 238.0508 amu;$
Mass of hydrogen atom is

$M_{H}=\ 1.0078 amu$ ;
Mass of neutron

$m_{N}=$

$1.0087 amu$ ; atomic number of Uranium

$Z$

$=$ 92; Mass Number of Uranium

$A$

$=$ 238)

0%
$7.580$
0%
$6.216$
0%
$5.162$
0%
$3.146$

Explanation

No. of protons

$=$

$92$

No. of neutrons

$= 238 - 92 = 146$

$\Delta m = [92\times 1.0078 + 146\times 1.0087 - 238.0508] amu$

$= 1.937 amu$

$BE = 1.937\times 931.478 MeV$

$= 1804.27 MeV$

$BE/per\ nucleon = \dfrac{1804.27}{238}MeV$

$= 7.58MeV$

The energy required to split

$^{16}_{8}O$ nucleus into four

$\alpha$ particles is

$\underline{\hspace{0.5in}}$ .
(The mass of an

$\alpha$ particle is

$4.0026.03 amu$ and that of oxygen is

$15.994915 amu$ )

0%
$11.6 MeV$
0%
$13.7 MeV$
0%
$14.4 MeV$
0%
$15.8 MeV$

Explanation

$^{16}_8O \rightarrow 4 ^4_2{He}$

$\Delta m = [4\times 4.002603 - 15.994915] \mu$

$= 0.015497 \mu$

$Q = \Delta m C^2$

$= 0.015497\times 931.478\ MeV$

$= 14.44 \ MeV$

The kinetic energy of the

$\alpha$ - particle emitted in the decay

$^{238}_{94}Pu \rightarrow ^{234}_{92}U +$

$^4_2He$ .
(The atomic masses of

$^{238}Pu, ^{234}U$ and

$\alpha$ particle are 238.04955u, 234.04095 u and 4.002603u respectively. Neglect any recoil of the nucleus)

0%
4.392 MeV
0%
6.259 MeV
0%
5.592 MeV
0%
4.952 MeV

Explanation

Assuming that energy librated by mass defect is converted into KE of

$\alpha -$ particle.

$\Delta m = (238.04955 - 234.04095 - 4.002603) \mu$

$= 0.005997 \mu$

$KE = E = 0.005997\times 931.478\ MeV$

$KE = {5.586} \ MeV$

The energy released in the following process is :

$A + B$

$\rightarrow$

$C + D +Q$
(mass of

$A$ is

$1.002$ amu ; mass of

$B$ is

$1.004$ amu ;mass of

$C$ is

$1.001$ amu ; mass of

$D$ is

$1.003$ amu)

0%
1.254 MeV
0%
0.931 MeV
0%
0.465 MeV
0%
1.862 MeV

Explanation

$\Delta m = [1.002 + 1.004 - 1.001 - 1.003]$

$= 0.002 \mu$

$E = Q = 0.002\times 931.478 MeV$

$= 1.862 MeV$

In nuclei with mass number greater than

$20$ , the average binding energy is:

0%
$8MeV$
0%
$0.8MeV$
0%
$80MeV$
0%
$0.08MeV$

Explanation

The binding energy per nucleon Vs number of nucleon curve suggests that average binding is almost

$8\ MeV$ for the nuclei having mass number greater than

$20$ .

The masses of neutron and proton are

$1.0087$ and

$1.0073$ amu respectively. If the neutrons and protons combine to form a Helium nucleus of mass

$4.0015 amu$ , the binding energy of the Helium nucleus will be :

0%
$28.4 MeV$
0%
$20.8 MeV$
0%
$27.3 MeV$
0%
$14.2 MeV$

Explanation

Mass of neutron

$= 1.0087 amu$

Mass of proton

$= 1.0073 amu$

In a helium nucleus,

protons

$= 2$

neutrons

$= 2$

Mass(theoritically)

$= (2\times1.0087 + 2\times1.0073)$

$=4.032$

Mass defect

$= (4.032- 4.0015)\ amu$

$=0.0305\ amu$

So,

$1\ amu$ gives

$931.5\ MeV$ of energy.

So,

$0.0305\times931.5 MeV$ of binding energy

$= 28.4 \ MeV$

$_{3}Li^{7}+_{1}H^{2}\rightarrow _{4}Be^{8}+_{o}n^{1}+Q$
Mass of

$_{3}Li^{7}=$

$7.01823amu$
Mass of

$_{1}H^{2}=$

$2.01474amu$
Mass of

$_{4}Be^{8}=$

$8.00785 amu$
Mass of

$_{o}n^{1}=$

$1.00899 amu$
Then, the value of Q is

0%
$5\ MeV$
0%
$10\ MeV$
0%
$15\ MeV$
0%
$0$

Explanation

Mass of reactants is:

$7.01824 + 2.01474 = 9.03297\ amu$
Mass of products is:

$8.00785 + 1.00899 = 9.01684\ amu$
The difference in mass

$= 0.01613\ amu$ . (This mass has been converted to energy).

$1 amu = 1.67 \times 10^{-27} kg$
Hence, mass difference is

$= 0.01613 \times 1.67 \times 10^{-27} kg = 2.69371 \times 10^{-29} kg$
Using Einstein's relation:

$E = mc^{2}$

$E = 2.69371 \times 10^{-21} \times (3 \times 10^{8})^{2} = 2.42 \times 10^{-12} J$

$E = 15\ MeV$

The mass defect and binding energy of

$^{12}_{6}$ C nucleus is
(Mass of

$^{12}_{6}$ C

$=$ 12.000000 amu;

$m_{p}=$ 1.007825 amu and

$m_{n}=$ 1.008665 amu)

0%
$0.06522$ amu; $72.1$ MeV
0%
$0.09894$ amu; $92.1$ MeV
0%
$0.05315$ amu; $102.2$ MeV
0%
$0.05315$ amu; $82.2$ MeV

Explanation

$=$ 6 neutrons + 6 protons

$\Delta m = (6\times 1.007825 + 6\times 1.008665 - 12)$

$= 0.09894 \ amu$

$E = (0.09894\times 931.478 \dfrac{MeV}{C^2})C^2$

$= 92.1 MeV$

The binding energy per nucleon of

$C^{12}$ is

$7.68$ MeV and that of

$C^{13}$ is

$7.47$ Mev. The energy required to remove one neutron from

$C^{13}$ is

0%
$495\ MeV$
0%
$49.5\ MeV$
0%
$4.95\ MeV$
0%
$0.495\ MeV$

Explanation

$C^{13} \rightarrow C^{12}_1 + ^1n$

${BE)_C}^{12} = 12\times 7.68$

$= 92.16\ MeV$

${BE)_C}^{13} = 13\times 7.47$

$= 97.11\ MeV$

$Q = {BE)_C}^{12} - {BE)_C}^{12}$

$= 97.11 - 92.16$

$={4.95\ MeV}$

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

0%
$11.56 \times 10^{-4}$
0%
$23.5 \times 10^{-4}$
0%
$33.5 \times 10^{-4}$
0%
$47.15 \times 10^{-4}$

Explanation

Deutron

$=\ ^{2}_{1}H$ i.e.,

$1$ proton and

$1$ neutron.

Packing fraction

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

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

$= 11.56 \times10^{-4}$

The atomic mass of

$_{7}N^{15}$ is

$15.000108 amu$ and that of

$_{8}O^{16}$ is

$15.994915 amu$ . The minimum energy required to remove the least tightly bound proton is (mass of proton is

$1.007825 amu)$

0%
$0.01 3018 amu$
0%
$12.13 MeV$
0%
$13.018 MeV$
0%
$12.13 eV$

Explanation

$_8^{16}O \rightarrow ^{15}_7N + ^1_1P$

$\Delta m = [15.000108 + 1.007825 - 15.994915]$

$= 0.013018 amu$

$\Delta E = 0.013018\times 931.478$ MeV

$={12.13\ MeV}$

$2g$ of hydrogen is converted into

$1.986 gm$ Helium in a thermonuclear reaction, the energy released is

0%
63 X 10 $^{7}J$
0%
63 X 10 $^{10}J$
0%
126 X 10 $^{10}J$
0%
6.3 X 10 $^{20}J$

Explanation

$\Delta m = 2 - 1.986$

$= {0.014 gm}$

$\Delta E = (\dfrac{0.014}{1000}Kg)\times (3\times 10^8)^2$

$= 126\times 10^{10} J$

$_{7}N^{14} +_{2}He^{4}\rightarrow$ X+

$_{1}H^{1}$ ; X is

0%
$_{9}F^{18}$
0%
$_{10}Ne^{18}$
0%
$_{8}O^{17}$
0%
$_{6}C^{15}$

Explanation

$_7N^{14} + _2He^4 \rightarrow _Z^A{X} + _1H^1$

Change balance:

$7 + 2 = Z + 1$

$Z = 8$
Mass balance:

$14 + 4 = A + 1$

$A = 17$

Bombardment of lithium with proton gives rise to the following reaction :

$^{7}_{3}Li+^{1}_{1} H \rightarrow$ 2(

$^{4}_{2}He) + Q$
The Q-value is
(atomic masses of lithium, proton and helium are

$7.016$ amu,

$1.008$ amu and

$4.004$ amu respectively)

0%
$14.904$ MeV
0%
$16.774$ MeV
0%
$10.634$ MeV
0%
$18.633$ MeV

Explanation

$^7_3{Li} + ^1_1H \rightarrow 2 ^4_2{He} + Q$

Mass defect

$= (7.016 + 1.008 - 2\times 4.004)\mu$

$= 0.016 amu$

$1amu = 931.478 \dfrac{MeV}{C^2}$

$E = 0.016\times 931.478 \dfrac{MeV}{C^2}\times C^2$

$=14.903 MeV$

$4.6\times 10^{22}$ atoms of an element weight

$13.8$ g. What is the atomic mass of the element?

0%
290 u
0%
180.6 u
0%
34.4 u
0%
104 u

Explanation

$1$ 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.

$\gamma$ -ray photon creates an electron-positron pair. The rest mass equivalent of electron is 0.5 MeV. KE of the electron - positron system is 0.78 MeV. Then the energy of

$\gamma$ -ray photon is

0%
$0.28 MeV$
0%
$1.78 MeV$
0%
$1.28 MeV$
0%
$0.14 MeV$

Explanation

The rest mass of electron is

$= 0.5\ MeV$

As, the momentum is conserved the rest mass of positron should also be

$0.5\ MeV$ .

The kinetic energy of the pair is

$0.78\ MeV$ .

$\therefore$ The energy of

$\gamma$ ray photon is

$= (0.5+0.5+0.78) \ MeV= 1.78 \ MeV$

Assertion (A) : All the radioactive elements are ultimately converted into lead
Reason (R): All the elements above lead are unstable

0%
Both A and R are true and R is correct explanation of A
0%
Both A and R are true and R is not correct explanation of A
0%
A is true but R is false
0%
A is false but R is true

Explanation

For all

$4n, 4n+2$ series elements, on emission of different kind of rays, they form different isotopes of lead and also all element above lead are unstable .
But,

$4n+1$ series elements (Neptunium series), end in

$^{209}_{83}Bi$ (Bismuth not lead)

Assertion (A): Binding energy per nucleon is the measure of the stability of the nucleus.

Reason (R): Binding energy per nucleon is more for heavier nuclides.

0%
Both A and R are true and R is correct explanation of A.
0%
Both A and R are true and R is not the correct explanation of A.
0%
A is true but R is false
0%
A is false but R is true

In the nuclear fusion reaction :

$^{2}_{1}H+ ^{3}_{1}H \rightarrow$

$^{4}_{2}He+n$ , given that the repulsive potential energy between the two nuclei is

$\sim 7.7 \times 10^{-14}$ J, the temperature to which the gases must be heated to initiate the reaction is nearly
(Boltzmann's constant

$k=$ 1.38 x 10

$^{-23}$ J)

0%
$10^{7} K$
0%
$10^{5} K$
0%
$10^{3} K$
0%
$10^{9}K$

Explanation

The repulsive potential energy

$=\dfrac{3}{2}kT$

$7.7\times 10^{-14}=\dfrac{3}{2}\times 1.38\times\times 10^{-23} \times T$

$\rightarrow \ T = 3.7\times 10^9 K \sim 10^9 K$

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.

0%
Both A and R are true and R is correct explanation of A
0%
Both A and R are true and R is not correct explanation of A
0%
A is true but R is false
0%
A is false but R is true

A nuclear transformation is denoted by X(n,

$\alpha$ )

$\rightarrow _{3} Li^{7}$ . The nucleus of element X is

0%
$_{6}C^{12}$
0%
$_{5}B^{10}$
0%
$_{5}B^{9}$
0%
$_{4}Be^{11}$

Explanation

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

$_{5}B^{10}$

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.

0%
Both A and R are true and R is correct explanation of A
0%
Both A and R are true and R is not correct explanation of A
0%
A is true but R is false
0%
A is false but R is true

Assertion (A): Nuclear fusion reactions are considered as thermo-nuclear reactions

Reason (R): The source of stellar energy is nuclear fusion

0%
Both A & R are true and R is the correct explanation of A
0%
Both A & R are true and R is not correct explanation of A
0%
A is true but R is false
0%
A is false but R is true

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

0%
A and B are correct
0%
A and B are false
0%
A is true, B is false
0%
A is false, B is true

Explanation

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.

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

0%
$\dfrac{2\sqrt{2}}{3}c$
0%
$\dfrac{2}{3}c$
0%
$\dfrac{1}{3}c$
0%
$\dfrac{1}{4}c$

Explanation

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:

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

An electron moves with a speed of

$\dfrac{\sqrt{3}}{2}c$ . Then its mass becomes_____ times its rest mass. (Given velocity of light

$=c$ )

0%
$2$
0%
$3$
0%
$3/2$
0%
$4$

Explanation

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.

To obtain an isotope of a given radioactive atom, the atom must emit

0%
one alpha and one beta particle
0%
one alpha and two beta particle
0%
two alpha and two beta particle
0%
three alpha and four beta particle

Explanation

Emission of-particle decreases the atomic number of the product by two and each

$\beta$ - emission increases the atomic number by one. So the atomic number remains the same and the mass number is different. after emitting one alpha and 2 beta.

The atomic mass of

$_{8}O^{16}$ is

$15.9949 amu$ . Mass of one neutron and one proton is

$2.016490 amu$ and the mass of an electron is

$0.00055 amu$ . The binding energy per nucleon of oxygen atom is

0%
$7.97 MeV$
0%
$11.5 MeV$
0%
$22.8 MeV$
0%
$82.3 MeV$

Explanation

The number of proton

$p=8$ and number of neutron,

$n=16-8=8$

Total number of nucleon

$N=16$

Combined mass of nuclei

$=$ total mass of proton + total mass of neutron

$=8\times 2.016490=16.13192$ amu

Mass defect

$\Delta m =16.13192-15.9949=0.13702$ amu

Binding energy per nucleon

$=\Delta m c^2/N=(0.137\times 931.49)/16= 127.6141/16=7.975MeV$ where

$1 amu =931.49/c^2 MeV$

The mass of a

$^7_3Li$ nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of

$^7_3Li$ nucleus is nearly

0%
46 MeV
0%
5.4 MeV
0%
3.9 MeV
0%
23 MeV

Explanation

$\begin{aligned}\Delta m &=0.042 \\&=0.042+2\times 1.66 \times 10^{-27}\mathrm{~kg} \\\text { Binding energy }&=\Delta \mathrm{m} c^{2} \\\text { Binding energy per nucleon } \\=& \frac{\Delta m c^{2}}{7} \\&[\operatorname{mass number} \text { of } l i=7]\end{aligned}$

$\begin{array}{l}=\frac{0.042 \times 1.66\times 10^{-27} \times\left(3 \times 10^{8}\right)^{2}}{7} \mathrm{~J} \\\text { To convert the energy in ev from } \\\text { joule, we will divide it by } 1.6 \times 10^{-19} \\=\frac{0.042 \times 1.66 \times 10^{-27} \times\left(3 \times 10^{8}\right)^{2}}{7 \times 1.6 \times 10^{-19}}\mathrm{ev} \\=5.4 \times 10^{6}\mathrm{ev} \\\therefore \text { Energy }\operatorname{lost} =5.4\mathrm{Mev}\end{array}$

The mass of proton is 1.0073 u and that of neutron is 1.0087 u (u

$=$ atomic mass unit). The binding energy of

$^{4}_{2}He$ is (Given : helium nucleus mass

$\approx$ 4.0015 u)

0%
0.0305 J
0%
0.0305 erg
0%
28.4 MeV
0%
0.061 u

Explanation

$BE=\Delta m\times 931$

$=[2(1.0087+1.0073)-4.0015]\times 931$

$=28.4 MeV$

Which of the following is true for the following isotopes of uranium

U

$^{235}$ and U

$^{238}$ ?

0%
both contain the same number of neutrons
0%
both contain the same number of protons, electrons and neutrons
0%
both contain the same number of protons and electrons but U $^{238}$ contains three more neutrons than U $^{235}$
0%
U $^{238}$ contain three less neutrons than U $^{235}$

Explanation

Natural Uranium contains

$3$ radioactive isotopes

${ U }^{ 234 },{ U }^{ 235 }$ and

${ U }^{ 238 }$

${ U }^{ 238 }$ has a mass number

(proton

$92+$ neutron

$146$ )

$=238$

${ U }^{ 235 }$ has mass number (proton

$92+$ neutron

$143$ )

$=235$

Hence we can say that both contain the same number of protons and electrons but

${ U }^{ 238 }$ contains three more neutrons than

${ U }^{ 235 }$

In the process of nuclear fusion

0%
only heavy nucleus break into light nuclei
0%
fusion of light nuclei at normal temperature takes place
0%
fusion of light nuclei at high pressure and low temperature takes place
0%
fusion of light nuclei at high pressure and high temperature takes place

When four hydrogen nuclei fuse together to form a helium nucleus, then in this process

0%
energy is absorbed.
0%
energy is liberated.
0%
absorption and liberation of energy depends upon the temperature.
0%
energy is neither liberated nor absorbed.

The binding energy per nucleon in deutorium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a
helium nucleus the energy released in the fusion is -

0%
2.2 MeV
0%
28.0 MeV
0%
30.2 MeV
0%
23.6 MeV

For a nuclear fusion process, suitable nuclei are

0%
Any Nuclei
0%
Heavy Nuclei
0%
Light Nuclei
0%
Nuclei lying in the middle of Periodic table

The heavier nuclei tend to have larger N/Z ratio because

0%
a neutron is heavier than a proton
0%
a neutron is an unstable particle
0%
a neutron does not exert electric repulsion
0%
Coulomb force has longer range as compared to the nuclear force

Atoms of same element having same atomic number but different mass number is

0%
Isobars
0%
Isotone
0%
Isotope
0%
None of these

Explanation

$\begin{array}{l}\text { Atoms having same atomic number but different mass } \\\text { number is called Isotope. } \\\therefore \text { Isotope is correct }\end{array}$

Principle of Hydrogen bomb is .................. reactions.

0%
Uncontrolled fusion
0%
Fission
0%
Controlled fusion
0%
None of these

One a.m.u. or one 'u' is equal to :

0%
$1.66053892 10^{-27} kg$
0%
$1.66053892 10^{-29} kg$
0%
$1.23457656 10^{-27} kg$
0%
$1.23457656 10^{-29} kg$

Explanation

One a.m.u.(atomic mass unit) or one 'u' is

$\dfrac{1}{12}$ of the mass of one carbon-

$12$ atom.
It is equal

$1.6605389210 \times 10^{-24} g$ or

$1.6605389210 \times 10^{-27} kg$ .

The modern atomic mass unit is based on the mass of :

0%
C-12 isotope
0%
hydrogen
0%
oxygen
0%
nitrogen

Explanation

One a.m.u. or one 'u' is equal to

$1.66053892 10^{-24}\ g$ or

$1.66053892 10^{-27}\ kg$ . It is equal to

$\dfrac{1}{12}$ of the mass of an atom of carbon-12.

It is used as a standard. The masses of all other atoms are determined relative to the mass of an atom of carbon-12.

The binding energies energy per nucleon for

$C^{12}$ is

$7.68\ MeV$ and that for

$C^{13}$ is

$7.5\ MeV.$ The energy required to remove a neutron from

$C^{13}$ is

0%
$5.34\ MeV$
0%
$5.5\ MeV$
0%
$9.5\ MeV$
0%
$9.34\ MeV$

Explanation

Total binding energy of

$C^{13}$ is

$E_1=7.5\times 13=97.5\ MeV$

Total binding energy of

$C^{12}$ is

$E_2=7.68\times 12=92.16\ MeV$

The energy required to remove a neutron is

$E_1-E_2=5.34\ MeV$

What is the mass of one atom of

$C-12$ in grams?

0%
1.992 $\displaystyle \times 10^{-23}$ gm
0%
1.989 $\displaystyle \times 10^{-23}$ gm
0%
1.892 $\displaystyle \times 10^{-23}$ gm
0%
1.965 $\displaystyle \times 10^{-23}$ gm

Explanation

Mass of

$1$ mole =

$12\,gm$

Mass of

$6.022 \times 10^{23}\,atom = 12\,gm$

mass of

$1$ atom =

$\dfrac{12}{6.023 \times 10^{23}} = 1.993 \times 10^{-23}\,gm$

Find the uncertainty in mass.

0%
$1.4\times10^{-35}$ $kg$
0%
$2.6\times10^{-11}$ $kg$
0%
$2.6\times10^{-13}$ $kg$
0%
$8\times10^{-14}$ $kg$

Explanation

$\triangle E.\triangle t$ =

$h/2\pi$

$\triangle E$

$=\displaystyle\ \frac{h}{2\pi \triangle t}$ =

$\displaystyle\ \frac {6.62\times10^{-34}}{2\pi\times8.4\times10^{-17}}$

$=1.24\times10^{-18}$

$\triangle mc^{2}$ =

$\triangle E$

Uncertainty in mass is

$\triangle m$

$=\displaystyle\ \frac{1.24\times10^{-18}}{9\times10^{16}}$

$=1.38\times10^{-35}\approx 1.4\times10^{-35}$

$kg$

Isotopes are the atoms of the same element which contain equal number of

0%
nucleons
0%
neutrons
0%
protons
0%
neutrons and protons

Find the binding energy of

$_{28}^{62}\textrm{Ni}$ , Given

$m_{H}$ =

$1.008$

$u$ ,

$m_{n}$ =

$1.0087$

$u$ ,

$_{28}^{62}\textrm{m}$ =

$62.9237$

$u$

0%
$545.3$ $MeV$
0%
$595.3$ $MeV$
0%
$645.3$ $MeV$
0%
$695.3$ $MeV$

Explanation

Number of protons in

$_{28}^{62}Ni$ ,

$p=28$

Number of neutrons in

$_{28}^{62}Ni$ ,

$n=28$

Given:

$m_{H}=1.008u , m_{n}=1.0087u$ ,

$_{28}^{62}m=61.9237u$

Then mass defect,

$\Delta m=28m_{H}+34m_{n}-_{28}^{34}m$

$\Delta m=28\times1.008+34\times1.0087-61.9237$

$\Delta m=0.5961u$

Hence binding energy,

$B.E.=\Delta m\times931$

$B.E.=0.5961\times930=554.3\ MeV$

The kinetic energy of

$\alpha$ -particles at a distance

$5 \times 10^{-14}$ m from the nucleus will be(in Joules)

0%
$6.4 \times 10^{-13}$
0%
$4.3 \times 10^{-13}$
0%
$2.1 \times 10^{-13}$
0%
$3.4 \times 10^{-14}$

Explanation

The charge on an alpha particle is

$+2$ .

Hence

$q = 2e = 3.2 \times 10^{-19} C$

This alpha particle is accelerated by a potential of

$2 \times 10^{6} V$

Hence Energy of the particle

$= qV = 3.2 \times 10^{-19} \times 2 \times 10^{6} = 6.4 \times 10^{-13} J$

This is the total energy.

Potential energy when it is at a distance of

$5 \times 10^{-14} m$ from the nucleus

$PE = \dfrac{1}{4\pi \epsilon_{0}}\dfrac{q_{1}q_{2}}{d}$

Nucleus contains 47 protons.

Hence,

$q_{2} = 47e = 47 \times 1.6 \times 10^{-19} C$

Substituting these values:

We get PE =

$4.3 \times 10^{-13} J$

$KE + PE = TE$

$KE = TE - PE = (6.4 - 4.3) \times 10^{-13} J = 2.1 \times 10^{-13}J$

Find the fraction of the mass.

0%
$5.28\times10^{-6}$
0%
$5. 84\times10^{-7}$
0%
$5.28\times10^{-8}$
0%
$7.8\times10^{-9}$

Explanation

Given :

$\Delta t = 8.4\times 10^{-17}$ s

As we know the relation

$\triangle E.\triangle t$

$=\dfrac{h}{2\pi}$

$\therefore$

$\triangle E$

$=\displaystyle\ \frac{h}{2\pi \triangle t}$ =

$\displaystyle\ \frac {6.62\times10^{-34}}{2\pi\times8.4\times10^{-17}}=1.24\times10^{-18}$ J

From mass-energy equivalance relation :

$\triangle mc^{2}$ =

$\triangle E$
Uncertainty in mass

$\triangle m$

$= \dfrac{\Delta E}{c^2} =\displaystyle\ \frac{1.24\times10^{-18}}{(3\times10^{8})^2}=1.38\times10^{-35}\approx 1.4\times10^{-35}$

$kg$
Mass of neutral pion

$m_{pion} = 264 m_{electron} = 264\times 9.1\times 10^{-31}$ kg

$\therefore$ Fraction of mass

$\displaystyle\ \frac{\triangle m}{m_{pion}}$ =

$\displaystyle\ \frac{1.38\times10^{-35}}{264\times9\times10^{-31}}=5.28\times10^{-8}$

The average

$\ KE$ of molecules in a gas at temperature

$\ T$ is

$\displaystyle \frac {3}{2}kT$ . Find the temperature at which the average KE of molecules equal to binding energy of its atoms.

0%
$\ 1.05 \times 10^{5}K$
0%
$\ 1.05 \times 10^{4}K$
0%
$\ 1.05 \times 10^{3}K$
0%
none of these

Explanation

Binding energy of atoms

$=13.6\ eV=2.1789\times 10^{-18}J$

Now since this energy is equal to

$\dfrac{3}{2}kT$

$T=\dfrac{2}{3}\dfrac{2.1789\times10^{-18}}{k}=1.05\times 10^5K$

As the mass number

$A$ increases, the binding energy per nucleon in a nucleus

0%
increases
0%
decreases.
0%
remains the same.
0%
varies in a way that depends on the actual value of $A$ .

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 4 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 4 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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