MCQ Questions for Class 12 Medical Physics Nuclei Quiz 7

One atomic mass unit is equal to?

(Given, mass of one carbon-12 atom

$= 1.992 \times 10^{-23}$ )

0%
$1.66\times 10^{-27}g$
0%
$1.66\times 10^{-24}g$
0%
$1.66\times 10^{-23}g$
0%
$1.66\times 10^{-25}g$

Explanation

One atomic mass unit (amu or u) is equal to 1/12th of the mass of an atom of carbon-12.

Mass of an atom of carbon-12

$= 1.992 \times 10^{-23}\ g$

1 amu

$=$ 1/12 mass of carbon-12 atom

$= \dfrac {1}{12} \times 1.992 \times 10^{-23}$

$= 1.66\times 10^{-24}g$

Find out the missing particle for the nuclear reaction as shown below:

$\displaystyle _{92}^{238}{U} \rightarrow _{90}^{234}{Th} + (?)$

0%
$\alpha$
0%
$_{-1}e^0$
0%
neutron
0%
photon
0%
proton

Explanation

The given reaction :

$^{238}_{92} U \rightarrow ^{234}_{90} Th$

$+$

$^A_ZX$

In nuclear reactions, the mass number as well as the atomic number is conserved.

$\therefore$

$238 = 234 + A$

$\implies A =4$

Also,

$92= 90+Z$

$\implies Z = 2$

Thus the missing particle is

$^4_2He$ (alpha particle).

When two light nuclei fuse to form a relatively heavier nucleus, the specific binding energy of the product nucleus is:

0%
lower than that of the reacting nuclei
0%
equal to that of the reacting nuclei
0%
greater than that of the reacting nuclei
0%
equal to exactly half of either of the reacting nuclei

Choose the decay that ejects the heaviest particle?

0%
$\alpha$ - decay
0%
${\beta}^{-}$ decay
0%
${\beta}^{+}$ decay
0%
Electron capture
0%
$\gamma$ - decay

Explanation

Among electron, positron, alpha particle and gamma ray, the alpha particle is the heaviest particle having mass

$4$ times that of proton and is ejected in the Alpha decay reaction.

Alpha decay :

$^A_ZX \rightarrow ^{A-4}_{Z-2}Y + ^4_2He$ where

$^4_2He$ is the alpha particle

Radioactive changes and ordinary chemical changes are different, because radioactive changes:

0%
are explosive
0%
gain energy
0%
happens in the nucleus
0%
release energy

$_1^2{H}+_1^3{H}\rightarrow _0^1{n}+$ X
The product X in the nuclear reaction represented is.

0%
$_1^1{H}$
0%
$_2^3{He}$
0%
$_2^4{He}$
0%
$_3^4{Li}$
0%
$_3^5{Li}$

Explanation

$^{4}_{2}{He}$

Fusion of deuterium and tritium creating helium - 4, freezing a neutron and releasing 17.59MeV as kinetic energy of products while a corresponding amount of mass disappears, in agreement with kinetic

$E=\Delta mc^{2}$ where

$\Delta m$ is the decrease in the total restass of particles.

The nuclear fusion reaction between deuterium and tritium takes place

0%
at ordinary temperature and pressure
0%
at low temperature and low pressure
0%
at very high temperature and very high pressure
0%
when the temperature is near absolute zero

The mass defect in a nucleus is

$3.5$ amu. Then the binding energy of the nucleus is.

0%
$3258.5$ MeV
0%
$325.85$ MeV
0%
$32.58$ MeV
0%
$3.258$ MeV

Explanation

Binding energy =(mass defect)

$c^{2} = (\Delta m) c^{2}$

$=\Delta(931)MeV$

$=3.5×931 = 3258.5 MeV$

Which of the following radioactive particle are decayed in the reaction

$^{218}_{85}At \rightarrow ^{214}_{83}Bi$ +

$^A_Z X$ .

0%
alpha
0%
$\beta^-$
0%
$\beta^+$
0%
electron capture
0%
gamma

Explanation

The given reaction:

$^{218}_{85} At \rightarrow ^{214}_{83} Bi$

$+$

$^A_Z X$

In a nuclear reaction, mass number and the atomic number are conserved.

$\therefore$

$218 = 214 + A$

$\implies A = 4$

Also,

$85 = 83+Z$

$\implies Z = 2$

Thus, the decayed particle is

$^4_2 He$ (alpha particle).

$_{ 1 }^{ 2 }{ H }+_{ 1 }^{ 2 }{ H }\quad \rightarrow \quad _{ 2 }^{ 3 }{ He }+X$
From the above nuclear fusion reaction, Identify the particle

$X$ .

0%
Positron
0%
Electron
0%
Proton
0%
Neutron
0%
Alpha particle

Explanation

Let

$A$ be the mass number of X and Z be the atomic number of it.

From mass number conservation,

$2+2=3+A$ or

$A=1$

From atomic number conservation,

$1+1=2+Z$ or

$Z=0$

$Z=0$ means neutral so X will be neutron.

The two alpha particles are released along with energy, when Li-7 is bombarded with a proton. It was found that the mass of the two alpha particles weighs less that the original product in the reaction. Now the mass that was converted to energy is called as :

0%
Einstein conversion
0%
mass defect
0%
theory of relativity
0%
natural transmutation
0%
chain reaction

A chlorine atom whose nucleus contains

$17$ protons and

$20$ neutrons. Find out the composition of the nucleus of an isotope of chlorine?

0%
$20$ protons, $17$ neutrons
0%
$19$ protons, $18$ neutrons
0%
$18$ protons, $18$ neutrons
0%
$17$ protons, $19$ neutrons
0%
$16$ protons, $20$ neutrons

Which of the following particles may differ in numbers in neutral isotopes of the same element?

0%
protons
0%
neutrons
0%
electrons
0%
alpha particles
0%
positrons

Explanation

Isotopes of the same element have same number of protons but different number of neutrons.

For example :

$^{35}_{17} Cl$ and

$^{37}_{17} Cl$ are isotopes of each other.

Both have

$17$ protons but first

$Cl$ has

$18$ neutrons and other

$Cl$ has

$20$ neutrons.

Six protons and six neutrons are brought together to form a carbon nucleus, but the mass of the carbon nucleus is less than the sum of the masses of the individual particles due to the mass defect, This deducted mass has been

0%
converted into the binding energy of the nucleus.
0%
given of in a radioactive decay process.
0%
converted into electrons.
0%
converted into energy to hold the electrons in orbit
0%
emitted as light.

Explanation

Nuclear binding energy accounts for a noticeable difference between the actual mass of an atom's nucleus and its expected mass based on the sum of the masses of its non-bound components. The release in energy accounts for the stability of the bound atom.
Quantitatively, mass defect=

$\Delta M=M_{protons}+M_{neutrons}-M_{atom}$

Identify which of the following equations represents alpha decay in a nuclear fission reaction ?

0%
$_{ 1 }^{ 2 }{ H }+_{ 1 }^{ 2 }{ H }\rightarrow \quad _{ 1 }^{ 3 }{ H }+_{ 1 }^{ 1 }{ H }+$ energy
0%
$_{ 88 }^{ 226 }{ Ra }\rightarrow \quad _{ 86 }^{ 222 }{ Rn }+_{ 2 }^{ 4 }{ He }$
0%
$_{ 83 }^{ 209 }{ Bi }\rightarrow \quad _{ 84 }^{ 209 }{ Po }+_{ -1 }^{ 0 }{ e }$
0%
$_{ 92 }^{ 235 }{ U }+_{ 0 }^{ 1 }{ n }\rightarrow \quad _{ 54 }^{ 140 }{ Xe }+_{ 38 }^{ 94 }{ Sr }+_{ 2 }^{ 1 }{ n }$
0%
$_{ 2 }^{ 4 }{ He }+_{ 13 }^{ 27 }{ Al }\rightarrow \quad _{ 15 }^{ 30 }{ P }+_{ 0 }^{ 1 }{ n }$

Explanation

In alpha decay reactions, alpha particle

$(^4_2He)$ is emitted while the atomic number and mass number of parent nucleus decreases by

$2$ and

$4$ respectively.

Thus reaction in option B is an alpha decay reaction.

Isotopes of same element exists in nature.How they differ from each other ?

0%
They will have different locations on the periodic table.
0%
They will have different numbers of electrons.
0%
They will have undergo different chemical reactions.
0%
They have different numbers of protons.
0%
They have different numbers of neutrons.

Which of the following sometimes requires initiation from a neutron?

0%
nuclear fission
0%
radiation therapy
0%
radioactive tracer
0%
radioactive carbon dating

When

$_{92}^{238}U$ undergoes alpha decay, what is the daughter nucleus?

0%
$_{90}^{236}Th$
0%
$_{91}^{236}Pa$
0%
$_{90}^{234}Th$
0%
$_{94}^{242}Pu$
0%
$_{94}^{240}Pu$

Explanation

Let the daughter nucleus be

$^A_ZX$ .

Alpha decay of Uranium :

$^{238}_{92}$

$\rightarrow$

$^{A}_ZX$

$+$

$^4_2 He$

Conservation of mass number :

$238 = A+ 4$

$\implies A = 234$

Conservation of atomic number :

$92 = Z+2$

$\implies Z =90$

Thus the daughter nucleus is

$^{234}_{90} Th$ .

On bombardment of neutron with boron,

$\alpha$ -particle is emitted. The product nuclei formed is:

0%
$_{6}C^{12}$
0%
$_{8}Li^{6}$
0%
$_{4}Be^{9}$
0%
$_{3}Li^{7}$

Explanation

$_{3}Li^{7}$

$^{10}B +n(0.025eV)\rightarrow_{4}He^{2}+_{7}Li^{3}+2.79MeV$

Lithium is formed on the bombardedment of neutron with boron by emitting

$\alpha$ particles.

What is the molar mass of ammonium carbonate

${({NH}_{4})}_{2}{CO}_{3}$ ?

0%
$48\ g/mol$
0%
$96\ g/mol$
0%
$82\ g/mol$
0%
$78\ g/mol$
0%
$192\ g/mol$

Explanation

$(NH_4)_2CO_3$ is the chemical formula of ammonium carbonate.

$N = 14 \times (2) = 28$

$H = 1 \times (4 \times 2) = 8$

$C = 12 \times 1 = 12$

$O = 16 \times 3 = 48$

Molar mass

$= 28 + 8 + 12 + 48 = 96\ g/mol$

How many grams are there in a

$7.0$ mole sample of sodium hydroxide?

0%
$40.0g$
0%
$140.0\ g$
0%
$280.0\ g$
0%
$340.0\ g$
0%
$420.0\ g$

Explanation

The formula of sodium hydroxide is

$NaOH$

Atomic weight of

$Na$ =

$23\ {g}/{mol}$

Atomic weight of

$O$ =

$16\ {g}/{mol}$

Atomic weight of

$H$ =

$1\ {g}/{mol}$

$\therefore$ molecular weight of

$NaOH$ = atomic mass of

$Na$ + atomic mass of

$O$ + atomic mass of

$H$

$23 + 16 + 1 = 40\ {g}/{mol}$

Therefore, 1 mole of

$NaOH = 40\ g/mol$

7 moles of

$NaOH = 7 \ mole \times 40\ g/mol= 280g$

Hence, option

$B$ is correct.

Calculate the mass defect.

0%
$2.014$ u
0%
$1.0087$ u
0%
$0.01876$ u
0%
$0.0234$ u

Explanation

Possible fusion reaction

$_{ 1 }^{ 2 }{ H }+_{ 1 }^{ 3 }{ H }\rightarrow _{ x }^{ y }{ He }+_{ 0 }^{ 1 }{ n }\\ 1+1=x+0\Rightarrow x=2\\ 2+3=y+1\Rightarrow y=4\\ _{ 1 }^{ 2 }{ H }+_{ 1 }^{ 3 }{ H }\rightarrow _{ 2 }^{ 4 }{ He }+_{ 0 }^{ 1 }{ n }$

Mass defect

$=\triangle m=\left( m\left( _{ 1 }^{ 2 }{ H } \right) +m\left( _{ 1 }^{ 3 }{ H } \right) \right) -\left( m\left( _{ 2 }^{ 4 }{ He } \right) +m\left( _{ 0 }^{ 1 }{ n } \right) \right) \\ =(2.014012+3.016050-4.002603-1.0087)u\\ =0.018759u\approx 0.01876u$

Which of the following isotopes is used for treatment of cancer?

0%
$Co^{60}$
0%
$K^{40}$
0%
$Sr^{90}$
0%
$I^{131}$

Explanation

$Co^{60}$

Cobalt therapy or cobalt -60 therapy is the medical use of gamma rays from the radioisotope cobalt 60 to treat conditions such as cancer.

Isotopes of an element differ in the.

0%
atomic number
0%
Number of nucleons
0%
Number of neutrons
0%
Mass number

The Q value is the ________ energy released in the decay at rest.

0%
kinetic
0%
potential
0%
thermal
0%
optical

Explanation

The Q value is the amount of energy released by a nuclear reaction. It is based on the mass-energy equivalence.

The Q value is defined as change in kinetic energy, i.e

$Q=KE_{final}-KE_{initial}=(m_{final}-m_{initial})c^2$ .

Which of the following is a nuclear reaction?

0%
Two hydrogen atoms combine to form hydrogen molecule
0%
Sodium atom gives up an electron to become sodium ion
0%
Water splits up into hydrogen and oxygen by electrolysis
0%
Isotopes of Hydrogen nuclei combine to form helium nuclei

Explanation

Isotopes of hydrogen nuclei combine to form helium nuclei

$_{1}H^{3}+_{1}H^{2}\rightarrow_{2}He^{4}+n^{1}$ + energy

$4_1H^1\rightarrow \ _2He^4+2e^0_{+1}+26$ MeV represents

0%
Fusion
0%
Fission
0%
$\beta$ -decay
0%
$\gamma$ decay

The equation

$E=mc^2$ was theoretical. It received experimental proof from.

0%
Nuclear reactions carried out in the laboratory
0%
Measurement of molecular mass of a compound formed by chemical comination
0%
Measurement of molecular mass of products of decomposition
0%
The phenomenon of radioactivity

Explanation

The phenomenon of radioactivity

The equation

$E=m{c}^{2}$ was found therotically and it is not proved initially. Later in the theorem received the proof from the phenomenon of radioactivity.

The difference between a nuclear reactor and an atomic bomb is that

0%
no chain reaction takes place in nuclear reactor while in the atomic bomb there is a chain reaction
0%
the chain reaction in nuclear reactor is controlled
0%
the chain reaction in nuclear reactor is not controlled
0%
no-chain reaction takes place in atomic bomb while it takes place in nuclear reactor

A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2 :The ratio of their nuclear size will be

0%
$2^{1/3} : 1$
0%
$1 : 3^{1/2}$
0%
$3^{1/2} : 1$
0%
$1:2^{1/3}$

Explanation

Since the linear momentum of the system must be conserved, the total momentum finally after disintegration remains same as that in beginning

$=0$

$\implies m_1v_1-m_2v_2=0$

$\implies \dfrac{m_1}{m_2}=\dfrac{v_2}{v_1}=\dfrac{1}{2}$

Since the density of both particles is same,

$\dfrac{\rho\dfrac{4}{3}\pi r_1^3}{\rho\dfrac{4}{3}\pi r_2^3}=\dfrac{1}{2}$

$\implies \dfrac{r_1}{r_2}=\dfrac{1}{2^{1/3}}$

Assume that the nuclear binding energy per nuclear

$(B/A)$ versus mass number

$(A)$ as shown in the figure. Use this plot to choose the correct choice (s) given below.

0%
Fusion of two nuclei with mass number lying in the range of $51 < A<100$ will release energy
0%
Fission of a nucleus lying in the mass range of $200 < A < 260$ will release energy when broken into two equal fragments
0%
Both (a) and (b)
0%
None of the above

The binding energy/nucleon of deuteron

$(_{1}H^{2})$ and the helium atom

$(_{2}He^{4})$ are

$1.1$ MeV and

$7$ MeV respectively. If the two deuteron atoms fuse to form a single helium atom, then the energy released is.

0%
$26.9$ MeV
0%
$25.8$ MeV
0%
$23.6$ MeV
0%
$12.9$ MeV

Explanation

The nuclear reaction is:

$_1H^2 + _1H^2 \rightarrow _2He^4$

Binding energy of deutron,

$E_d = 1.1\times 2 = 2.2$ MeV

Binding energy of helium,

$E_{He} = 7\times 4 = 28$ MeV

Energy released,

$Q = E_{He} - 2E_d$

$\therefore$

$Q = 28 - 2(2.2) = 23.6$ MeV

An alpha particle (

$^4He$ )has a mass of 4.00300 amu. A proton has mass of 1.00783 amu and a neutron has mass of 1.00867 amu respectively. The binding energy of alpha particle estimated from these data is the closest to

0%
27.9 MeV
0%
22.3 MeV
0%
35.0 MeV
0%
20.4 MeV

Explanation

An alpha particle

$(_2^4He)$ has 2 protons and two neutrons.

Mass defect

$\Delta M = 2m_p + 2m_n - m_{He}$

$\therefore$

$\Delta M=2(1.00783) + 2(1.00867)-4.00300$

$= 0.03000$ amu

Binding energy of alpha particle B.E

$= \Delta M c^2$

$\therefore$ B.E

$= 0.03000 \times 931.5 =27.9$ MeV

$(\because 1$ amu

$c^2=931.5 MeV$

$)$

When a radioactive isotope

$_{88} Ra^{228}$ decays in series by the emission of three

$\alpha$ -particles and a

$\beta$ particle the isotope finally formed is :

0%
$_{84} X^{220}$
0%
$_{86}X^{222}$
0%
$_{83}X^{216}$
0%
$_{83}X^{215}$

Explanation

In each alpha decay, mass number of the parent nucleus decreases by

$4$ units while atomic number decreases by

$2$ units. In each beta decay, atomic number increases by

$1$ unit but mass number remains the same.
So, mass number of the isotope formed

$A = 228-3\times 4 = 216$
Atomic number

$Z = 88-3\times 2+1 = 83$
So correct answer is option C.

Radiocarbon is produced in the atmosphere as a result of

0%
collision between fast neutrons and nitrogen nuclei present in the atmosphere
0%
action of ultraviolet light from the sun on atmospheric oxygen
0%
action of solar radiations particularly cosmic rays on carbon dioxide present in the atmosphere
0%
lightning discharge in atmosphere

Explanation

Radiocarbon is produced in the atmosphere as result of collision between fast neutrons and nitrogen nuclei present in the atmosphere.
Nuclear reaction is given as :

$_7N^{14} \ + \ _0n^1 \ \rightarrow \ _6C^{14} \ + \ _1H^1$

Isotopes of an element contain

0%
the same number of protons but different number of neutrons
0%
the same number of neutrons but different number of protons
0%
equal number of protons and electrons
0%
equal number of nucleons

Explanation

Isotopes of an element must have same atomic number

$(Z)$ but different mass number

$A$ .

Number of protons is equal to the atomic number.

So, isotopes of an element have same number of protons.

Mass number is equal to the sum of number of protons and neutrons i.e.

$A = p+n$

As isotopes of an element have different mass number but same number of protons, thus they must have different number of neutrons.

The explosion of hydrogen bomb is based on the principle of

0%
Uncontrolled fission reaction
0%
Nuclear fusion reaction
0%
Controlled fission reaction
0%
Photoelectric effect

What is the binding energy per nucleon of

$_{6}C^{12}$ nucleus?
Given : mass of

$C^{12}(m_c)=12.000\ u$
mass of proton

$9M_p)=1.0078\ u$
mass of neutron

$(m_n)=1.0087\ u$ and

$1\ amu=931.4\ MeV$

0%
$5.26\ MeV$
0%
$10.11\ MeV$
0%
$15.65\ MeV$
0%
$7.68\ MeV$

Consider the following statements.
(i) All isotopes of an element have the same number of neutrons.
(ii) Only one isotope of an element can be stable and non-radioactive.
(iii) All elements have isotopes.
(iv) All isotopes of Carbon can form chemical compounds with Oxygen-

$16$ .
The correct option regarding an isotope is

$?$

0%
(iii) and (iv) only
0%
(ii), (iii) and (iv) only
0%
(i), (ii) and (iii) only
0%
(i), (iii) and (iv) only

A nucleus

$X$ initially at rest, undergoes alpha decay according to the equation

${92}^{{X}^{A}} \rightarrow {Z}^{{Y}^{228}} + \alpha$
Then, the values of

$A$ and

$Z$ are

0%
$94,230$
0%
$232,90$
0%
$190,32$
0%
$230,94$

Explanation

Here,

$\alpha=$

$_2He^4$

Thus,

$_{92}X^A \rightarrow _ZY^{228}+$

$_2He^4$

For balance the equation,

$A=228+4=232$ and

$92=Z+2$ or

$Z=90$

Thus, option B will be the right option.

The masses of neutron and proton are 1.0087 and 1.0073 amu respectively. If the neutrons and protons combine to form 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

Total protron in helium is 2 and neutron in helium is 2

mass defect

$\Delta M=2(1.0087+1.0073)-4.0015=0.0305 amu$

Binding energy

$E=931\Delta M=28.4MeV$

An atom of

$_{53}^{131}I$ and an atom of

$_{53}^{127}I$ contain the same number of:

0%
Quarks
0%
Neutrons
0%
Nucleons
0%
Protons

Explanation

Atomic representation of an element is

$^A_ZX$

where:

$X:$ Atomic Symbol

$Z:$ Atomic number

$A:$ Mass number

Hence,

$^{131}_{53}I$ and

$^{127}_{53}I$ has same atomic number but different mass number. Hence, they have same number of protons but different number of neutrons.

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).

0%
$2.5$ MeV
0%
$8.0$ MeV
0%
$0.5$ MeV
0%
$7.5$ MeV

Explanation

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

$\Delta E = E_1 - E_2$

$\therefore$

$\Delta E = 82.5 -80 = 2.5 MeV$

The mass defect of

$_{ 2 }^{ 4 }{ He }$ is

$0.03 u$ . The binding energy per nucleon of helium (in

$MeV$ ) is

0%
$69.825$
0%
$6.9825$
0%
$2.793$
0%
$27.93$

Explanation

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

$\dfrac{E}{n} = \dfrac{27.93}{4} =6.9825$

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

0%
$19 : 81$
0%
$10 : 11$
0%
$15 : 16$
0%
$81 : 19$

Explanation

Let the percentage of

$B^{10}$ atoms be

$x$ , then average atomic weight

$= \dfrac {10x + 11(100 - x)}{100} = 10.81$

$\Rightarrow x = 19$

$\therefore \dfrac {N_{B^{10}}}{N_{B^{11}}} = \dfrac {19}{81}$ .

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

$= 185\ MeV$ ).

0%
$43.5\ MW$
0%
$58.5\ MW$
0%
$69..6\ MW$
0%
$73.1\ MW$

Explanation

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

$= 58.47\times 10^{6}W$

$\equiv 58.5\ MW$ .

The equation,

$4_1H^1 \rightarrow [ _2He^4]^{2+} + 2e^- ; + 26 MeV$ represents :

0%
$\gamma -$ decay
0%
$\beta -$ decay
0%
Fisslon
0%
Fusion

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.

0%
$94,230$
0%
$232,90$
0%
$190,32$
0%
$230,94$

Explanation

The decay equation is

$_{92}X^A\rightarrow _ZY^{228}+\alpha$

$\alpha$ -particle is nucleus of

$_2He^4$

$_{92}X^A\rightarrow _ZY^{228}+_2He^4$

$A=228+4=232$

$Z=92-2=90$ .

Nucleus of mass number

$A$ , originally at rest, emits the

$\alpha$ particle with speed

$v$ . The daughter nucleus recoils with a speed of :

0%
$2v(A+4)$
0%
$4v(A+4)$
0%
$4v/(A-4)$
0%
$2v/(A-4)$

Explanation

According to conservation of momentum

$4v=(a-4)v'\Rightarrow v'=\cfrac { 4v }{ A-4 }$

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

0%
$36.2\ MeV$
0%
$23.6\ MeV$
0%
$47.2\ MeV$
0%
$11.8\ MeV$
0%
$9.31\ MeV$

Explanation

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,

$= 28 - 2 \times 2.2 = 28 - 44 = 23.6\ MeV$ .

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 7 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 7 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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