MCQ Questions for Class 12 Medical Physics Nuclei Quiz 9

$U-238$ nucleus originally at rest, decays by emitting an

$\alpha-$ particle, say with a velocity of

$v\ m/s$ . The recoil velocity in

$( ms^{-1})$ of the residual nucleus is

0%
$\dfrac {4v}{238}$
0%
$-\dfrac {4v}{238}$
0%
$\dfrac {v}{4}$
0%
$-\dfrac {4v}{234}$

Explanation

$\textbf{Step 1: Applying momentum conservation}$

Let

$v_r$ be the velocity of the residual nucleus.

Since there are not any external forces acting on the system, therefore momentum is conserved.

$P_f =P_i$

$m_{\alpha}=4m$ ,

$v_{\alpha}=v$

$m_r=(238-4)m=234m$ , where

$m$ is mass of a proton

$P_i=0$ , as nucleus is at rest

$\Rightarrow$

$m_{\alpha}v_{\alpha} +m_r v_r=0............(1)$

$\textbf{Step 2: Solving Equation}$

Substituting the values in equation 1

$4mv+ 234mv_r=0$

$\Rightarrow v_r=-\dfrac{4v}{234}$

Hence, Option

$D$ is correct

The binding energy per nucleon of deuteron is

$1.2MeV$ and that of the helium atom is

$7.1MeV$ . What is the energy released, if two deuteron atoms combine to form a single helium atom?

0%
17.8MeV
0%
19.5MeV
0%
21,2MeV
0%
23.6MeV

Explanation

$\begin{array}{c}\text { Given - } * B E \text { per nucleon of deuteron }=1.2 \mathrm{MeV} \\\text { and of Helium }=7.1\mathrm{MeV}\\{ }_{1}^{2} \mathrm{H}+{}_{1}^{2}\mathrm{H}\longrightarrow_{4}^{2}\mathrm{He}\end{array}$

$\begin{array}{l}\Rightarrow \text { Energy released = BE of Helium -2.BE of Deutron. } \\\text { BE }=B E \text { ber nucleons } X \text { Num of nucleons. } \\\begin{aligned}\Rightarrow E &=[4 \times 7.1]-2\times[1.2 \times 2] \text { Mev } \\E &=23.6\mathrm{MeV}\end{aligned}\end{array}$

$1.0\ MeV$ neutron emitted in a fusion loses one half of its kinetic energy in each collision with a moderator molecule. Number of collision it must be more to reach thermal energy will be

$\left(Take, \dfrac {3}{2}k_{B}T=0.040\ eV\right)$

0%
$5$
0%
$25$
0%
$125$
0%
$625$

Explanation

$\begin{array}{l}\text { Let "n" be required number } \\\text { of collision } \\\text { Given } \quad \frac{3}{2} k_{B} T=0.04 \\(0.04) n=1 \\\qquad n=25\end{array}$

$X(n, \alpha)+_0^1n \rightarrow _{3}^{7}Li + _2^4He$ , then

$X$ will be :

0%
$_{5}^{10}B$
0%
$_{5}^{9}B$
0%
$_{4}^{11}Be$
0%
$_{2}^{4}He$

Explanation

$\underbrace { x\left( n,d \right) }_{ _{ a }^{ b }{ x } } +\dfrac { 1 }{ 0 } n\rightarrow \dfrac { 7 }{ 3 } li+\dfrac { 4 }{ 2 } He$ ......(given reaction)

$\displaystyle \sum$ Atomic number in

$L.H.S=a+0=a$

$\displaystyle \sum$ Atomic number in

$R.H.S=3+2=5$

$\therefore \ a=5$

$\displaystyle \sum$ Atomic mass number in

$R.H.S=7-4=11$

$\displaystyle \sum$ Atomic mass number in

$L.H.S=b+1$

$\therefore \ b+1=11$

$\therefore \ 6=10$

$\therefore \ x\to _{ a }^{ b }{ x } \to _{ 5 }^{ 10 }{ x }$

comparing this with options we get

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

Hence option

$A$ is correct.

The binding energy of deutron is

$2.2 \space MeV$ and that of

$^4_2 He$ is

$28 \space MeV$ . If two deutrons are fused to form one

$^4_2 He$ then the energy released is:

0%
$25.8 \space MeV$
0%
$23.6 \space MeV$
0%
$19.2 \space MeV$
0%
$30.2 \space MeV$

Explanation

$_1{D^2}{ \to _2}H{e^4}$

Energy released

$=28 - 2 \times 2.2 = 23.6\,MeV$

Hence,

option

$(B)$ is correct answer.

$x \ g$ of

$A$ (atomic mass

$50$ ) contains

$n$ atoms, how many atoms are there in

$20 \ x \ g \ B$ (atomic weight

$100$ )

0%
$n$
0%
$10 n$
0%
$20 n$
0%
$\dfrac{n}{10}$

$m_{0}$ is the mass of isotope

$\frac{16}{8}O, m_{p}\ and\ m_{m}$ are the masses of a proton and neutron respectively, the nuclear binding energy of the isotope is:

0%
$(m_{0}\ -\ 17m_{n})c^{2}$
0%
$(m_{0}\ -\ 8m_{p})c^{2}$
0%
$(m_{0}\ -\ 8m_{p}-8m_{n})c^{2}$
0%
$m_{0}c^{2}$

Explanation

$\Delta m = \text{Mass defect=mass of constituent nucleons - mass of molecule}=8m_p +(16-8)m_n -m_0$

Binding energy =

$\Delta m \times c^2=(8(m_p+m_n)-m_0 )c^2$

Option C is correct.

Calculate the neutron separation energy from the following data.

$m(^{40}_{20}Ca) = 39.962591 \ u;$

$m (^{41}_{20}Ca) = 40.962278 \ u;$

$m_n = 1.00865;$

$1u = 931.5 MeV/C^2$

0%
$7.57 \space MeV$
0%
$8.36 \space MeV$
0%
$9.12 \space MeV$
0%
$9.56 \space MeV$

Explanation

Energy equation:

$_{ 20 }^{ 41 }{ Ca }+Q-1\rightarrow_{ 20 }^{ 40 }{ Ca }+_{ 0 }^{ 1 }{ n}$

Mass defect,

$\Delta m=m(_{ 20 }^{ 40 }{ Ca })+m(_{ 0 }^{ 1 }{ n })-m(_{ 20 }^{ 41 }{ Ca }\\ \quad=39.962591+1.008665-40.962273\\ \quad=0.008978u$

BUt

$1u=931.5MeV.C^2$

Hence

$0.008978 u=0.008978\times931.5=8.363MeV$

Thus separation energy

$=8.363MeV$

Which of the following graph shows variation of decay rate of a radioactive sample with number of nuclei left in the sample :-

Explanation

Decay rate

$r=-\dfrac{dN}{dt}$ and

$r=N\lambda$

Where

$\lambda(decay-constant)$ is a constant so

$r=\dfrac{-dN}{dt} \propto N$

Which is a linear relation and

$negative$ sign denote that as N decrease the rate

$-\dfrac{dN}{dt}$ increase but

$\dfrac{dN}{dt} decreases$

So option D is correct.(Line with negative slope)

A slow neutron strikes a nucleus of

$^{235}_{92} {U}$ splitting it into lighter nuclei of

$^{141}_{56} {Ba}$ and

$^{92}_{36} {Kr}$ along with three neutrons. The energy released in this reaction is :
(The masses of Uranium, Barium and Krypton in this reaction are

${235.043933}$ a.m.u,

${140.917700}$ a.m.u and

${91.895400}$ a.m.u respectively. The mass of a neutron is

$1.008665$ a.m.u)

0%
$198.9$ MeV
0%
$156.9$ MeV
0%
$186.9$ MeV
0%
$209.8$ MeV

Explanation

${}_{92}^{235}U + {}_0^1n \to {}_{56}^{141}Ba + {}_{36}^{92}Kr + 3({}_0^1n) + Q$

Q is Energy released Q = ( Total mass of reactants – total mass ) $\times {C^2}$

$= \left[ {Mass\left( {{}_{92}^{235}U} \right) + mass\left( {{}_0^1n} \right)} \right] - \left[ {mass{}_{56}^{141}Ba + mass{}_{36}^{92}Kr + mass3{}_0^1n} \right]{c^2}$

$= \left( {235 \cdot 043933 - 140.917700 - 91.8954 - 2 \times 1.008665} \right)u \times {c^2}$

$= 198.9MeV$

The following nuclear reaction is an example of

$\frac { 12 }{ 6 } C+\frac { 4 }{ 2 } H\rightarrow \frac { 16 }{ 8 } O+energy$ :

0%
Fission
0%
Fusion
0%
alpha decay
0%
beta decay

In hydrogen bomb we use the process of

0%
Fusion
0%
Fission
0%
Both (1) & (2)
0%
None of these

Explanation

The hydrogen bomb relies on fusion, the process of taking two separate atoms and putting them together to form a third atom.

Fusion is the process that powers the sun and the stars. It is the reaction in which two atoms of hydrogen combine together, or fuse, to form an atom of helium. In the process some of the mass of the hydrogen is converted into energy

Hence,

option

$A$ is correct answer.

The binding energies of two nuclei

$P^{n}$ and

$Q^{2n}$ are

$x$ and

$y$ joules. If

$2x > y$ then the energy released in the reaction:

$P^{n}+P^{n} \rightarrow Q^{2n}$ will be

0%
$2x+y$
0%
$2x-y$
0%
$-(2x-y)$
0%
$x+y$

Explanation

Given:

Binding energy of nuclei

$P^n=x$

Binding energy of nuclei

$Q^{2n}=y$

To find:

Energy released in the reaction.

$P^n+P^n\rightarrow Q^{2n}$

Total Binding energy of reactant =

$x+x$

$2x$

Total Binding energy of product =

$y$

Energy released in reaction =

$BE_{product}-BE_{reactants}$

$y-2x$

$-(2x-y)$

Thus, option (C) is correct.

$N_t=N_0e^{-\lambda t}$ then number of disintegrated atoms between

$t_1$ to

$t_2$ (

$t_2> t_1$ ) will be:

0%
$N_0[e^{\lambda t_2}-e^{\lambda t_1}]$
0%
$N_0[-e^{\lambda t_2}-e^{-\lambda t_1}]$
0%
$N_0[e^{-\lambda t_1}-e^{-\lambda t_2}]$
0%
none

Explanation

Initial amount is

$N_0$ so amount after time

$t_1$ will be

$N=N_0e^{-\lambda t_1}$

And that after time

$t_2$ is

$N_2=N_0e^{-\lambda t_2 }$

So decayed in period between

$t_1$ and

$t_2$ is nothing but

$N_1-N_2=N_0(e^{-\lambda t_1}-e^{-\lambda t_2})$ as

$N_1 \gt N_2$

option c is correct.

Binding energy per nucleon versus mass number curve for nuclei is shown in the figure. A, B, C and D are four muclei indicated on the curve. The process that would release energy is

0%
$C \rightarrow 2D$
0%
$A \rightarrow C+D$
0%
$A \rightarrow 2C$
0%
$B \rightarrow C +D$

Explanation

We can clearly deduce from the graph that:

(A)

$C > 2D$

(B)

$C > A\Rightarrow C+D > A$

(C)

$C > A\Rightarrow 2C > A$

(D)

$C > B\Rightarrow C+D > B$

Only in option A, the reactant side has more Binding energy than productant side thus releasing energy in the process.

Option (A) is correct.

1130841_1158467_ans_ee3fb244ff274297a22f91880a9fccd9.jpg

One mole of radium has an activity of 1/3.7 killo curie. Its decay constant will be

0%
$\dfrac{1}{6}\times 10^{-10}s^{-1}$
0%
$10^{-10}s^{-1}$
0%
$10^{-11}s^{-1}$
0%
$10^{-8}s^{-1}$

Explanation

Number of mole of radium, $n=1$

The number of nuclei of radium, $N={{N}_{a}}=avagdaro\,\,number=6.023\times {{10}^{23}}$

Activity, $A=\lambda N$

$\because 1\,\,curie=3.7\times {{10}^{10}}\,decay/\sec$

$\dfrac{1}{3.7}\,kilocurie=0.27\,kilocurie=9.99\times {{10}^{12}}\,dacay/\sec$

So, $\lambda =\dfrac{A}{N}=\dfrac{9.99\times {{10}^{12}}}{6.023\times {{10}^{23}}}=1.65\times {{10}^{-11}}\,$

The nuclear radius of a certain nucleus is

$7.2 \,fm$ and it has a charge of

$1.28 \times 10^{-17}C$ . The number of neutrons inside the nucleus is:

0%
126
0%
111
0%
118
0%
142

Explanation

Radius of nuclear given by,

$R = ROA^{1/3}$ where

$Ro = 1.25\, fm$

$\Rightarrow A = (\frac{R}{Ro})^{3} = (\frac{7.2}{1.25})^{3} = 191$ (atomic mass)

Now, no. of protons,

$Z = \frac{1.28\times 10^{-17}}{1.6\times 10^{-19}} = 80$

Thus, no.of neutrons nuclears:

$N = A-Z = 191-80 = 111$ (Ans)

1104823_1179146_ans_b928e314a812466cbe59136d5092c3f1.JPG

A particular hydrogen like atom has its ground state Binding Energy=122.4eV. It is in ground state. Then

0%
atomic number is 3
0%
an electron with 90 ev energy can interact with it and excite it
0%
An electron of kinetic energy 91.8eV can be brought to almost rest by this atom.
0%
An electron of KE nearly 2.6 eV may emerge from the atom electron of KE 125ev collides

Explanation

$\begin{array}{l}\text { Let Atomic number be "z" } \\\text { Binding energy }=z^{2}(13.6) \\\qquad \begin{array}{c}122.4=z^{2}(13 \cdot 6)\\z^{2}=9 \\z=3\end{array}\end{array}$

The kinetic energy of

$\alpha$ - particle emitted in the

$\alpha$ - decay of

${ _{ 88 }{ Ra }^{ 226 } }$ is [ given, mass number of Ra = 222 u ]

0%
5.201 Me V
0%
3.301 Me V
0%
6.023 MeV
0%
4.871 MeV

Explanation

$\begin{array}{l} _{ 88 }R{ a^{ 226 } }\to R_{ n }^{ 222 }+2H_{ e }^{ 4 } \\ Q=[226.02540-222.01750-4.00260]\times 931.5Mev \\ =0.0053\times 931.5=4.94Mev \\ K.E=\frac { { \left( { A-4 } \right) Q } }{ A } \\ =\frac { { 226-4 } }{ { 226 } } \times 4.94 \\ =4.85\, \, Mev \\ Q=4.94\, \, Mev, \\ K.E=4.85\, \, Mev \end{array}$

What is the binding energy of an electron in the first orbit of

${ Li }^{ 2+ }$ is

0%
$13.6 eV$
0%
$122.4 eV$
0%
$-13.6 eV$
0%
$-122.4 ev$

Explanation

$\begin{array}{l} \text { We know, } \\ \text { Energy of electron in Bohr's theory is } \\ \end{array}$

$E=\left(\frac{-13 \cdot 6 z^{2}}{n^{2}}\right)ev$

$\begin{aligned} \text { for } & L{i}^{2+} \\ & z=3 \\ & n=1 \end{aligned}$

$\begin{array}{l} E=\frac{-13.6 \times 9}{1} \mathrm{ev} \\ E=-122.4 \mathrm{ev} \end{array}$

Assuming the energy released per fission of

$U^{235}$ is 200 me V , the energy released in the fission of 1 kg of

$U^{235}$ is energy

0%
$0.91\times10^{11}J$
0%
$0.91\times10^{13}J$
0%
$8.19\times10^{11}J$
0%
$8.19\times10^{13}J$

Explanation

Given,

Energy released per atom, $E=200\,MeV=200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}=3.2\times {{10}^{-11}}\,J$

Number of atom in $1\,kg$ uranium, $n=\dfrac{m{{N}_{A}}}{M}=\dfrac{1000\times 6.023\times {{10}^{23}}}{235}=2.56\times {{10}^{24}}$

Total Energy released = $nE=2.56\times {{10}^{24}}\times 3.2\times {{10}^{-11}}=8.19\times {{10}^{13}}J$

Hence, Total Energy released = $8.19\times {{10}^{13}}J$ .

The average binding energy per nucleon of a nucleus is of the order of

0%
$8 \,eV$
0%
$8 \,J$
0%
$8 \,keV$
0%
$8 \,MeV$

In a fission reaction

$^{236}_{92}U \rightarrow ^{117}X+^{117}Y+n+n$ . The binding energy per nucleon of

$X$ and

$Y$ are

$8.5 MeV$ whereas of

$^{236}U$ is

$7.6 MeV$ . The total energy liberated will be about

0%
$400 MeV$
0%
$200 MeV$
0%
$300 MeV$
0%
$200 keV$

Explanation

$\begin{array}{l}\text { Total no. of nucleus in " } X \text { " }\\\text { and "Y" would be } \\\qquad \begin{aligned}117+117=234 &\\\text { Total B.E of } X \text { and } Y=8.5 \times 234 \\=1989\text { MeV }\end{aligned}\end{array}$

$\begin{array}{rl}^{236}U \ have \ B E=7.6 \times 236 \\& =1793.6\mathrm{MeV} \\\triangle E & =1989-1793.6 \\\Delta E & =195.4\mathrm{MeV} \\\triangle E & \approx 200 \mathrm{Mev}\end{array}$

$\text{Approximately}$

Consider the nuclear reaction

$X^{200}\rightarrow A^{110}+B^{90}$ if the binding energy per nucleon for X, A and B is 7.4 MeV, 8.2 MeV and 8.2 Mev respectively, what is the energy released?

0%
200 MeV
0%
160 MeV
0%
110 MeV
0%
90 MeV

Explanation

Total energy of

$X=2\omega\times 7.4\ MeV$

Total energy of

$A=110\times 8.2\ MeV$

Total energy of

$B=90\times 78.2\ MMeV$

So,

$2\omega\times 7.4=(110\times 8.2)+(90\times 8.2)$

Hence

$-$ energy

$=-2\omega\times 7.4+(110\times 8.2)+(90\times 8.2)\\=160\ MeV$

The binding energy per nucleon of

$_{26}F^{56}$ nucleus is

0%
$8.8\ MeV$
0%
$88\ MeV$
0%
$493\ MeV$
0%
$413\ MeV$

Explanation

$\begin{array}{l}\text { The binding energy per } \\\text { nucleon of } Fe \text { nucleus is } 8.8 \mathrm{MeV} \\\text { which is highest. }\end{array}$

A radioactive isotope has a half-life of

$2$ years. How long will it approximately take the activity to reduce to

$3\%$ of its initial value?

0%
$4.8\ Years$
0%
$7\ Years$
0%
$10\ Years$
0%
$9.6\ Years$

Explanation

$\begin{array}{l}=N_{0}\left(\frac{1}{2}\right)^{n} \\\text { Here, }\text {N is final activity. } \\\text { No is initial activity. }\\\text { Given: } \frac{N}{N_{0}}=\frac{3}{100}\\\therefore\quad\frac{3}{100}=\left(\frac{1}{2}\right)^{n}\\2^{n}=33.33 \\n \simeq 5\end{array}$

$\begin{array}{l}\therefore \text { Activity reduces to } 3 \% \text { of initial } \\\text { value of } 5 \text { half lives that is } 10\text { years. }\end{array}$

Electrons are bombarded to excite hydrogen atoms and six spectral lines are observed. If

$E_{g}$ is the ground state energy of hydrogen, the minimum energy the bombarding electrons should possess is

0%
$8E_{g}/9$
0%
$15E_{g}/16$
0%
$35E_{g}/36$
0%
$48E_{g}/49$

In a nuclear fusion reaction, two nuclei,

$A$ &

$B$ , fuse to produce a nucleus

$C$ , releasing an atom of energy

$\Delta E$ in the process. If the mass defects on the three nuclie are

$\Delta M_{A}, \Delta M_{B}, \Delta M_{C}$ respectively, then which of the following relations holds? Here

$c$ is the speed of light:-

0%
$\Delta M_{A}+\Delta M_{B}=\Delta M_{C}-\Delta E/c^{2}$
0%
$\Delta M_{A}+\Delta M_{B}=\Delta M_{C}+\Delta E/c^{2}$
0%
$\Delta M_{A}-\Delta M_{B}=\Delta M_{C}-\Delta E/c^{2}$
0%
$\Delta M_{A}-\Delta M_{B}=\Delta M_{C}+\Delta E/c^{2}$

Explanation

$\begin{array}{l}\text { Mass of photon is Difference in } \\\text { Mass defects of products and Reactants } \\\therefore\left[\Delta M_{C}\right]-\left[\Delta M_{A}+\Delta M_{B}\right]=\frac{\Delta E^{D}}{C^{2}} \\\triangle M_{A}+\Delta M_B=\Delta M_{C}-\frac{\Delta E}{C^{2}}\end{array}$

The reactor produced a power of

$10^{10}\ MW$ with

$50$ % efficiency. Then the number of deutrons required to run the reactor for one year will be approximately:

$m\left( _{ 1 }^{ }{ { H }_{ }^{ 2 } } \right) =2.014\ u,m\left( p \right) =1.007\ u,m\left( n \right) =1.008\ u,m\left( _{ 2 }^{ }{ { H }e_{ }^{ 4 } } \right) =4.001\ u$

0%
$5\times 10^{35}$
0%
$5\times 10^{32}$
0%
$5\times 10^{38}$
0%
$5\times 10^{40}$

Complete the equation for the following fission process

$_{92}U^{235}+_{0}n^{1}\rightarrow _{38}Sr^{90}+...$

0%
$_{54}Xe^{143}+3_{0}n^{1}$
0%
$_{54}Xe^{145}$
0%
$_{57}Xe^{142}$
0%
$_{54}Xe^{142}+3_{0}n$

Explanation

Sum of mass number in

$reactant$ side is

$235+1=236$

and Sum of atomic number in

$reactant$ side is

$92+0=92$

so the product side sum of mass number should also be

$236$ and that of atomic number should be

$92$

$_{54}Xe^{143} +3 _0n^1$ is

suitable because

$143+3\times 1 +90=236$ and

$38+54=92$

Nucleus of an elements contains 9 protons. It's valency would be:

0%
1
0%
2
0%
3
0%
5

Choose correct statement(s)

0%
The density of nuclear matter is independent of the size of the nucleus
0%
The binding energy per nucleon, for nuclei of middle mass number, is about $8$ MeV
0%
A free neutron is unstable
0%
A free proton is stable

The energy of the reaction

$Li ^ { 7 } + p \longrightarrow _2He^4$ is (the binding energy per nucleon in

$Li^7$ and

$He^4$ nuclei are

$5.60$ and

$7.06\ MeV$ respectively.)

0%
$17.3\ MeV$
0%
$1.73\ MeV$
0%
$1.46\ MeV$
0%
$Depends\ on\ the\ binding\ energy\ of\ proton.$

Explanation

$\begin{aligned}\text { Energy of } r \times n &=B \cdot E_{p}-B \cdot E_{R} \\&=7.6 \times 4 \times 2-5.6 \times 7 \\&={17.3 \mathrm{ Mev}}\end{aligned}$

If the mass of

$proton = 1.008 a.m.u.$ and mass of

$neutron =1.009 a.m. u.$ , then binding energy per nucleon for

$_4 Be^9$

$(mass = 9.012 a mu )$ would be:

0%
$40.065 Me v$
0%
$60.44 Me V$
0%
$67.2 Me v$
0%
$6.72 Me v$

Explanation

mass defect is

$\Delta m=4m_p+(9-4)m_n-m_{Be}=(4\times 1.008+5\times 1.009-9.012)amu=0.065amu$

so Binding energy of

$_4Be^9$ will be

$\Delta m \times 931.5Mev=.065\times 931.5Mev=60.5475Mev$

or the binding energy per nucleon will be

$\dfrac{BE}{nucleon}=\dfrac{60.5475}{9}=6.7275Mev$

The binding energy per nucleon for the parent nucleus is

$E_1$ and that for the daughter nuclei is

$E_2$ . Then :

0%
$E_1= 2E$
0%
$E_1>E_2$
0%
$E_2>E_1$
0%
$E_2=E_1$

The binding energies of the nuclei A and B are

$E _ { a }$ and

$E _ { b }$ respectively. Three nuclei of the element B fuse to give one nucleus of element 'A' and an energy 'Q' is released. Then

$E _ { a } , E _ { b } , \underline { Q }$ are related

0%
$E _ { a } - 3 E _ { b } = Q$
0%
$3 E _ { b } - E _ { a } = Q$
0%
$E _ { a } + 3 E _ { b } = Q$
0%
$E _ { b } + 3 E _ { a } = Q$

Explanation

The binding energy of reactants is

$Q_r=3E_b$

The binding energy of product is

$Q_p=E_a$

the enrgy released will be

$Q=Q_r-Q_p=3E_b-E_a$

A nuclear transformation is denoted by

$X(n, \alpha) \ ^7_3 Li$ , then the element X will be :-

0%
$^{10}_5 B$
0%
$^9_5 B$
0%
$^{11}_4 Be$
0%
$^4_2 He$

Explanation

Since Nuclear Transformtion is given as

$^A_ZX+^1_0n\rightarrow ^4_2He+^7_3Li$

On Applying conservation of mass no. and charge no.

$A+1=4+7$ (mass No. Conservation)

$Z+0=2+3$ (charge No. Conservation)

$\Rightarrow A=10$ and

$Z=5$

$\therefore$ Element will be

$^{10}_5B$

Atomic mass of

$_{ 6 }^{ 13 }{ C }$ is

$13.00335$ amu and its mass number is

$13.0$ . If amu

$=931$ MeV, the binding energy of the neutrons present in the nucleus is

0%
$0.24$ MeV
0%
$1.44$ MeV
0%
$1.68$ MeV
0%
$3.12$ MeV

Explanation

binding energy is the product of mass defect and c square .

now 13.00335 - 13 = 0.00335u.

=>B.E = 931

$\times0.00335$ = 3.12 Mev

According to binding energy per nucleon versus mass number curve, which is not correct?

0%
Two light nuclei fuse to form medium sized nuclei.
0%
A heavy nucleus fission to form two medium sized nuclei.
0%
Two medium sized nucleus fuse to form a heavy nucleus.
0%
The peak of the curve corresponds the most stable nucleus.

A person needs glasses for driving is suffering from which defect in his vision

0%
Astarmagnrtism
0%
Myopia
0%
Presbyopia
0%
Hypermetropia

An atomic power nuclear reactor can deliver

$300$ MW. The energy released due to fission of each nucleus of uranium atom

$U^{235}$ is

$170$ MeV. The number of uranium atoms fissioned per hour will be

0%
$4\times 10^{22}$
0%
$30\times 10^{22}$
0%
$10\times 10^{20}$
0%
$5\times 10^{15}$

Binding energy of deuterium is

$2.23\ MeV$ , then its mass defect in a.m.u. is

0%
$-0.0024$
0%
$-0.0012$
0%
$0.0012$
0%
$0.0024$

Explanation

Mass defect

$\Delta m=\dfrac {2.23}{931}=0.0024 amu$ .

1 atomic mass unit is equal to:

0%
$\tfrac{1}{12}^{th}$ mass of a carbon-12 atom
0%
$1.66\times 10^{-24}\ g$
0%
$6.023\times 10^{-23}\ g$
0%
$6.023\times 10^{23}\ g$

Explanation

One atomic mass unit is a mass unit equal to exactly one-twelfth (1/12th) the mass of one atom of carbon-12.

So, we can write 1 atomic mass unit

$= \dfrac {1}{12}$ mass of carbon-12 atom

$=\dfrac {1}{12}$

$\times$

$\dfrac {12}{{6.023 \times 10^{23}}} g$

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

A nuclear reactor delivers power of 10W, find fuel consumed by the reactor per hour if its efficiency is

$20\%$ , Given

$c=3 \times 10^8 m/s$

0%
$2 \times 10^{-6}g/hr$
0%
$9 \times 10^{-12}g/hr$
0%
$8 \times 10^{-9}g/hr$
0%
$2 \times 10^{-9}g/hr$

Explanation

As fuel consumed by the react or with efficiency 20%. Then energy will be:

$E = \dfrac{100 \times 100}{20} = 50 \ J/s$

$E= mc^2, m= \dfrac{E}{c^2}$

$m= \dfrac{50}{ ( 3 \times 10^8)^2} kg/s = \dfrac{50 \times 3600 \times 10^3}{( 3 \times 10^8)^2} g / hr= 2 \times 10^{-9} g / hr$

In a hydrogen atom, the binding energy of the electron in the nth state is

$E_n$ ,then the frequency of revolution of the electron in the nth orbit is:

0%
$2E_n/nh$
0%
$2E_nn/h$
0%
$E_n/nh$
0%
$E_nn/h$

Explanation

We know that, Kinetic energy=Binding energy

$\cfrac { 1 }{ 2 } m{ v }^{ 2 }={ E }_{ n }----(i)\\ mvr=\cfrac { nh }{ 2\pi } ----(ii)$

Dividing (i) by (ii),

$\cfrac { v }{ 2\pi r } =\cfrac { 2{ E }_{ n } }{ nh }$

Therefore, the frequency of revolution of the electron in the nth orbit is

$=\cfrac { 2{ E }_{ n } }{ nh }$

$\overset { 14 }{ \underset { 7 }{ N } }$ if mass attributed to electron were doubled & the mass attributed to protons were halved, the atomic mass would become approximately:-

0%
Halved
0%
Doubled
0%
Reduced by 25%
0%
Remain same

A heavy nucleus

$X$ of mass number

$240$ and binding energy per nucleon

$7.6MeV$ is split into two fragments

$Y$ and

$Z$ of mass numbers

$110$ and

$130$ the binding energy of nucleons in

$Y$ is

$8.5MeV$ per nucleon

$.$ Calculate the energy

$Q$ released per fission in

$MeV.$

0%
240 $\mathrm { Mev }$
0%
216 $\mathrm { Mev }$
0%
120 $\mathrm { Mev }$
0%
108 $\mathrm { Mev }$

Explanation

The nuclear reaction is

${X^{240}} \to {Y^{110}} + {Z^{130}} + Q$

As per given data

$Q = 110 \times 8.5 + 130 \times 8.5 - 240 \times 7.6$

$= 240\left( {0.9} \right)MeV = 216MeV$

Hence,

option

$(B)$ is correct answer.

What are the value of

$a$ and

$b$ respectively in the reaction

${}_9^4Be + {}_2^4He \to {}_b^aX + {}_0^1n$

0%
$17,7$
0%
$7,11$
0%
$12,6$
0%
$6,12$

Explanation

$_{ 9 }^{ 4 }{ Be }+_{ 2 }^{ 4 }{ He }\longrightarrow _{ b }^{ a }{ X }+_{ 0 }^{ 1 }{ n }$

or,

$_{ 9 }^{ 4 }{ Be }+_{ 2 }^{ 4 }{ He }\longrightarrow _{ 11 }^{ 7 }{ X }+_{ 0 }^{ 1 }{ n }$

$a=7$

$b=11$ .

Nucleus A is converted into C through the following reactions,
A

$\rightarrow$ B+

$\alpha$
B

$\rightarrow$ C+2

$\beta$
then,

0%
A and B are isotopes
0%
A and C are isobars
0%
A and B are isobars
0%
A and C are isotopes

Which one of the following cannot be used as a moderator in a nuclear reactor?

0%
Water
0%
Heavy water
0%
Molten sodium
0%
Graphite

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 9 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 9 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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