Four physical quantities are listed in Column I. Their values are listed in Column II in a random order. Column I Column II p. Thermal energy of air molecules at room temperature (i) 0.02 eV q. Binding energy of heavy nuclei per nucleon (ii) 2 eV r. X-ray photon energy (iii) 10 keV s. Photon energy of visible light (iv) 7 MeV The correct matching the Column I and Column II is given by

$$p\rightarrow i, q\rightarrow iv, r\rightarrow iii, s\rightarrow ii$$

$$p\rightarrow i, q\rightarrow iii, r\rightarrow ii, s\rightarrow iv$$

$$p\rightarrow ii, q\rightarrow i, r\rightarrow iii, s\rightarrow iv$$

$$p\rightarrow ii, q\rightarrow iv, r\rightarrow i, s\rightarrow iii$$

proton and neutron both are stable

MCQ Questions for Class 12 Medical Physics Nuclei Quiz 12

Mark out the correct statement(s)

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in both fission and fusion processes, the mass of reactant nuclide is greater than the mass of product nuclide
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in fission process, BE per nucleon of reactant nuclide is less than the binding energy per nucleon of product nuclide
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in fusion process, BE per nucleon of reactant nuclide is less than the binding energy per nucleon of product nuclide
0%
in fusion process, BE per nucleon of reactant nuclide is greater than the binding energy per nucloen of product nuclide

Explanation

Mass difference

$(\Delta m)=$ mass of product nuclide

$(m_p) -$ mass of reactant nuclide

$(m_r)$

We know that energy releases in both process so

$\Delta m$ should be negative.

Thus,

$m_r>m_p$

BE per nucleon

$=$ mass defect / A where mass defect = combined mass of proton and neutron

$-$ mass of nuclide.

$m_r>m_p$ so mass defect is more for reactant.

A proton and a neutron are both shot at

$100 ms^{-1}$ toward a

$_6^{12}C$ nucleus. Which particle, if either, is more likely to be absorbed by the nucleus?

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The proton.
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The neutron.
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Both particles are about equally likely to be absorbed.
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Neither particle will be absorbed.

Explanation

$^{12}_6C$ captures a neutron to form relatively stable element

$^{13}_6C$ .

$^{12}_6C$

$+^1_0n\rightarrow$

$^{13}_6C$

This is more stable than the otherwise formed element

$^{12}_7N$

Instantaneous power developed at time t due to the decay of the radionuclide is

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$\left (q_0t-\frac {q_0}{\lambda}+\frac {q_0}{\lambda}e^{-\lambda t}\right )E_0$
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$\left (q_0t+\frac {q_0}{\lambda}-\frac {q_0}{\lambda}e^{-\lambda t}\right )E_0$
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$\left (q_0t+\frac {q_0}{\lambda}+\frac {q_0}{\lambda}e^{-\lambda t}\right )E_0$
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$\left (q_0t-\frac {q_0}{\lambda}-\frac {q_0}{\lambda}e^{-\lambda t}\right )E_0$

Explanation

The radioactive nuclide (X) is produced at rate

${ q }_{ 0 }t$ and decays at the rate

$\lambda \left[ X \right]$
Therefore,

$\dfrac { d\left[ X \right] }{ dt } ={ q }_{ 0 }-\lambda \left[ X \right]$
Given, Energy released per decay is

${ E }_{ 0 }$ .

$\Rightarrow$ Instantaneous power at any time t is

$\lambda \left[ X \right] { E }_{ 0 }$
Solving, above differential equation, we get:

$\left[ X \right] = { q }_{ 0 }\left( \dfrac { t }{ \lambda } -\dfrac { 1 }{ { \lambda }^{ 2 } } \left( 1-{ e }^{ -\lambda t } \right) \right)$
Therefore, Instantaneous Power is

${ q }_{ 0 }\left( t-\dfrac { 1 }{ { \lambda } } \left( 1-{ e }^{ -\lambda t } \right) \right) { E }_{ 0 }$

Why is a

$_2^4He$ nucleus more stable than a

$_3^4Li$ nucleus?

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The strong nuclear force is larger when the neutron-to-proton ratio is higher.
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The laws of nuclear physics forbid a nucleus from containing more protons than neutrons.
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Forces other than the strong nuclear force make the lithium nucleus less stable.
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None of the above.

Four physical quantities are listed in Column I. Their values are listed in Column II in a random order.

Column I	Column II
p. Thermal energy of air molecules at room temperature	(i) 0.02 eV
q. Binding energy of heavy nuclei per nucleon	(ii) 2 eV
r. X-ray photon energy	(iii) 10 keV
s. Photon energy of visible light	(iv) 7 MeV

The correct matching the Column I and Column II is given by

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$p\rightarrow i, q\rightarrow iv, r\rightarrow iii, s\rightarrow ii$
0%
$p\rightarrow i, q\rightarrow iii, r\rightarrow ii, s\rightarrow iv$
0%
$p\rightarrow ii, q\rightarrow i, r\rightarrow iii, s\rightarrow iv$
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$p\rightarrow ii, q\rightarrow iv, r\rightarrow i, s\rightarrow iii$

Explanation

p) : Thermal energy of the air molecules at room temperature

$E_{thermal} \approx KT = 1.38 \times 10^{-23} \times 300 J = 4.14 \times 10^{-21} J = 0.259 eV$ (approx)

q) : The graph of binding energy per nucleon vs mass number suggests that the binding energy per nucleon of heavy atom lies between

$7$ to

$8 MeV$

r) : X-Rays have wavelength in the range from

$0.01 nm$ to

$10 nm$ ,

$\implies$ These have energy range of

$\approx 100 eV$ to

$100 keV$

s) : Visible light have wavelength

$\approx 380 nm$ to

$780 nm$ which implies that these have energy range

$1.6 eV$ to

$3.3 eV$

The equation

$4_1^1H^{2}\rightarrow _2^4He^{2+}+2e^{+1}+26\ MeV$ represents

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$\beta-decay$
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$\gamma-decay$
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$fusion$
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$fission$

In the core of nuclear fusion reactor, the gas becomes plasma because of

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strong nuclear force acting between the deuterons
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coulomb force acting between the deuterons
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coulomb force acting between deuteron-electron pairs
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the high temperature maintained inside the reactor core

The correct statement is

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the nucleus $_3^6Li$ can emit an alpha particle
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the nucleus $_{84}^{210}Po$ can emit a proton
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deuteron and alpha particle can undergo complete fusion
0%
the nuclei $_{30}^{70}Zn$ and $_{34}^{82}Se$ can undergo complete fusion

Explanation

$_{ 3 }^{ 6 }{ Li } \longrightarrow _{ 2 }^{ 4 }{ He }\quad (\alpha -particle)+ _{ 1 }^{ 2 }{ H }$

$_{ 3 }^{ 6 }{ Li } \longrightarrow _{ 2 }^{ 4 }{ He }\quad (\alpha -particle)+ _{ 1 }^{ 2 }{ H }\\ { M }_{ _{ 3 }^{ 6 }{ Li } }= 6.015123u\quad \\ { m }_{ _{ 2 }^{ 4 }{ He } }+{ m }_{ _{ 1 }^{ 2 }{ H } }= 4.002603u+ 2.014102u= 6.016705u\\ \Rightarrow { M }_{ _{ 3 }^{ 6 }{ Li } }> { m }_{ _{ 2 }^{ 4 }{ He } }+{ m }_{ _{ 1 }^{ 2 }{ H } }$
Therefore, it can emit an alpha particle.

$_{ 84 }^{ 210 }{ Po } \longrightarrow _{ 1 }^{ 1 }{ H }+ _{ 1 }^{ 3 }{ H } + _{ 82 }^{ 206 }Pb\\ { M }_{ _{ 84 }^{ 210 }Po }= 209.982876u \\ { m }_{ _{ 82 }^{ 206 }{ Pb } }+{ m }_{ _{ 1 }^{ 3 }{ H } }+{ m }_{ _{ 1 }^{ 1 }{ H } }= 205.974455u+ 3.016050u+ 1.007825u= 209.99833u\\ \Rightarrow _{ 84 }^{ 210 }{ Po }< _{ 1 }^{ 1 }{ H }+ _{ 1 }^{ 3 }{ H }+ _{ 82 }^{ 206 }Pb$
Therefore, polonium cannot emit a proton.

3)
They cannot undergo fusion as proved in the first example.

$_{ 30 }^{ 70 }{ Zn }+ _{ 34 }^{ 82 }{ Se } \longrightarrow _{ 64 }^{ 152 }{ Gd }\\ { M }_{ _{ 64 }^{ 152 }{ Gd } }= 151.919803u \\ { m }_{ _{ 30 }^{ 70 }{ Zn } }+{ m }_{ _{ 34 }^{ 82 }{ Se } }= 69.925325u+ 81.916709u= 151.842034u\\ \Rightarrow _{ 30 }^{ 70 }{ Zn }+ _{ 34 }^{ 82 }{ Se }< _{ 64 }^{ 152 }{ Gd }$
Therefore, fusion is not possible.

A nucleus with mass number

$220$ initially at rest emits an

$\alpha \ particle$ . If the

$Q$ value of the reaction is

$5.5 MeV$ , calculate the kinetic energy of the

$\alpha\ particle$

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$4.4 MeV$
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$5.4 MeV$
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$5.6 MeV$
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$6.5 MeV$

Explanation

$M \rightarrow X+ \alpha$

$E = \cfrac{p^2}{2m}$
Let Total energy =

$T$

$T = E_{\alpha} + E_{X}$

$M = M_{\alpha} + M_{X}$

$M_{X}= 216$
Conservation of momentum

$p_{\alpha} = p_{X}$

$m_{\alpha} E_{alpha} = m_XE_X$

$E_{\alpha} = \cfrac{m_x}{M}T$

$E_{\alpha} = \cfrac{216}{220} 5.5 = 5.4 MeV$

Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct choice(s) given below:

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Fusion of two nuclei with mass numbers lying in the range of $1 < A < 50$ will release energy.
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Fusion of two nuclei with mass numbers lying in the range of $51 < A < 100$ will release energy.
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Fission of a nucleus lying in the mass range of $100 < A < 200$ will release energy when broken into two equal fragments.
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Fission of a nucleus lying in the mass range of $200 < A < 260$ will release energy when broken into two equal fragments.

Explanation

The fact that there is a higher binding energy in the region near

$A=100-200$ means that either the breakup of heavier nuclei (fission) or the combining of lighter nuclei (fusion) will yield nuclei which are more tightly bound (less mass per nucleon) if the atom/s formed lies in this region of stability.
For fusion two atoms of atomic mass no. between

$50-100$ will combine and produce a atomic number of

$100-200$ and will be stable than individual atoms by releasing the energy. Hence, option B is correct.
For fission atom of atomic mass No. more than

$200-260$ will form two atoms of equal atomic number between

$100-200$ and new atoms will be more stable than parent atom by releasing the energy.

The rest energy of an electron is

$0.511 MeV.$ The electron is accelerated from rest to a velocity

$0.5 c$ . The change in its energy will be

0%
$0.026 MeV$
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$0.051 MeV$
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$0.07 MeV$
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$0.105 MV$

Explanation

$\displaystyle mc^2=\dfrac{m_0}{\sqrt{1-\dfrac{v^2}{c^2}}}c^2=\frac{m_0}{\sqrt{1-\dfrac{(0.5c)^2}{c^2}}}c^2$

$\displaystyle =\frac{m_0}{\sqrt{1-0.25}}c^2=\frac{m_0}{\sqrt{0.75}}c^2=1.15m_0c^2$

Change in energy

$= 1.15\:m_0c^2- m_0c^2$

$=0.15\times 9.1 \times 10^{-31}\times(3\times 10^8)^2$

$=12.285\times 10^{-15}J$

$\displaystyle =\frac{12.285\times10^{-15}}{1.6\times10^{-13} }MeV$

$= 0.07678\,MeV$

Binding energy per nucleon vs. mass number curve for nuclei is shown in fig. W, X, Y and Z are four nuclei indicated on the curve. The process that would release energy is

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$Y\rightarrow 2Z$
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$W\rightarrow X+Z$
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$W\rightarrow 2Y$
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$X\rightarrow Y+Z$

Explanation

Binding energy of

$W$ ,

$|E_W| = 120\times7.5=900$ MeV

Binding energy of

$Y$ ,

$|E_Y| = 30\times8.5=255$ MeV

In process :

$W\rightarrow 2Y$

$Q$ value,

$Q = |E_W| - |E_Y| = 900-2(255) =390$ MeV

$Q$ value is positive, thus energy is released in process

$W\rightarrow 2Y$

Let

$m_p$ be the mass of proton,

$m_n$ the mass of a neutron,

$M_1$ the mass of a

$_{10}^{20}Ne$ nucleus, and

$M_2$ the mass of a

$_{20}^{40}Ca$ nucleus. Then

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$M_2=2M_1$
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$M_2 > 2M_1$
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$M_2 < 2M_1$
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$M_1 < 10(m_p+m_p)$

Explanation

Every nucleus by virtue of its nuclear force has a certain binding energy, which results in mass defect, and makes the nucleus stable, despite the repulsive electrostatic forces b.w protons.
Neon being highly stable gas,

$_{10} Ne ^{20}$

$= 10$ protons

$+\ 10$ neutrons
So,

$M_1 < 10 (m_n+ m_p)$ for the nucleus to be stable.
Also the nuclear binding energy per nucleon increases at lower atomic nos.
Hence, per nucleon binding energy will be higher for calcium, hence weight/nucleon lesser

$M_2 < 2M_1$

The binding energies per nucleon for a deuteron and an -particle are

${ x }_{ 1} ,{ x}_{ 2 }$ respectively. What will be the energy Q released in the reaction

$_{ 1 }^{ }{ { H }_{ }^{ 2 } }+_{ 1 }^{ }{ { H }_{ }^{ 2 } }\rightarrow _{ 2 }^{ }{ { He }_{ }^{ 4 } }+Q$

0%
$4\left( { x }_{ 1 }+{ x }_{ 2 } \right)$
0%
$4\left( { x }_{ 2 }-{ x }_{ 1 } \right)$
0%
$2\left( { x }_{ 1 }+{ x }_{ 2 } \right)$
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$2\left( { x }_{ 2 }-{ x }_{ 1 } \right)$

The K.E. of the emitted

$\alpha-particle$ in the decay of

$^{226}_{88}Ra$ (approximately)

0%
2.3 MeV
0%
4.85 MeV
0%
9.7 MeV
0%
14 MeV

Explanation

$\begin{array}{c} _{88}^{226} Ra \longrightarrow{ }_{2}^{4} He^{2+}+{ }_{86}^{222} Rn \end{array}$

$226.02540 \mu \quad 4.00260 \mu \quad 222.01750 \mu$

$\begin{aligned} \Delta m &=0.0053 \\ \text { Energy } &=0.0053 \times 931.5 \mathrm{MeV} \\ &=4.94 \\ K_{1} &=\frac{m_{2}}{m_{1}+m_{2}} \times K \\ &=\frac{4}{222+4} \times 4.94 \\ &=4.8495 \end{aligned}$

Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which of these is most promising based on Lawson criterion?

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Deuteron density=2.0 x ${ 10 }^{ 12 }{ cm }^{ -3 }$ ,confinement time=5.0 x ${ 10 }^{ -3 }$ s
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Deuteron density=8.0 x ${ 10 }^{ 14 }{ cm }^{ -3 }$ ,confinement time=9.0 x ${ 10 }^{ -1 }$ s
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Deuteron density=4.0 x ${ 10 }^{ 23 }{ cm }^{ -3 }$ ,confinement time=1.0 x ${ 10 }^{ -11 }$ s
0%
Deuteron density=1.0 x ${ 10 }^{ 24 }{ cm }^{ -3 }$ ,confinement time=4.0 x ${ 10 }^{ -12 }$ s

Explanation

According to Lawson, deuteron density (n) and confinement tome

$t_0$ must satisfy the criterion

$nt_0 > 5\times 10^{14} cm^{-3}s$

This condition is satisfied only by choice(b) for which

$nt_0=(8.0\times 10^{14})\times (9.0\times 10^{-1})=7.2\times 10^{14}cm^{-3}s$ . Hence the correct choice is

$(b)$ .

The Q-value for the

$\alpha-decay$ of

$^{226}_{88}Ra$ is (approximately)

0%
4.93 MeV
0%
2.46 MeV
0%
9.8 MeV
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14.7 MeV

Explanation

$\begin{array}{c} _{88}^{226} Ra \longrightarrow{ }_{2}^{4} He^{2+}+{ }_{86}^{222} Rn \end{array}$

$226.02540 \mu \quad 4.00260 \mu \quad 222.01750 \mu$

$\begin{aligned} \Delta m=& 0.0053 \\ Q \text { -value } &=\Delta m \times 931.5 \mathrm{MeV} \\ &=0.0053 \times 931.5 \\ &=4.93695 \end{aligned}$

Which of the following nuclear reactions is not possible?

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$_{ 6 }^{ 12 }{ C+ }_{ 6 }^{ 12 }{ C }\longrightarrow _{ 10 }^{ 20 }{ Ne+ }_{ 2 }^{ 4 }{ He }$
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$_{ 4 }^{ 9 }{ Be+ }_{ 1 }^{ 1 }{ H }\longrightarrow _{ 3 }^{ 6 }{ Li+ }_{ 2 }^{ 4 }{ He }$
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$_{ 5 }^{ 11 }{ Be+ }_{ 1 }^{ 1 }{ H }\longrightarrow _{ 4 }^{ 9 }{ Be+ }_{ 2 }^{ 4 }{ He }$
0%
$_{ 3 }^{ 7 }{ Li+ }_{ 2 }^{ 4 }{ He }\longrightarrow _{ 1 }^{ 1 }{ H+ }_{ 4 }^{ 10 }{ B }$

Explanation

The following laws hold true in any reaction
1) Conservation of mass (nucleons)
2) conservation of charge
Checking for all the reactions

$\Sigma A_{products} = \Sigma A_{reactants}$
For option C:

$9 +4 \ne 11+1$
Hence, option (c) not possible.

The velocity of a body of rest mass

$m_o$ is

$\dfrac{\sqrt 3}{2} c$ (where c is the velocity of light in vacuum). The mass of this body is :

0%
$\left ( \dfrac{\sqrt3}{2} \right )m_o$
0%
$\left ( \dfrac{1}{2} \right )m_o$
0%
$3m_o$
0%
$2m_o$

Explanation

By special theory of relativity

$m' = \gamma m= \cfrac{m_0}{\sqrt{1 -\cfrac{v^2}{c^2}}}$

$m' = \cfrac{m_0}{\sqrt{1 -\cfrac{3}{4}}} =\cfrac{m_0}{(1/2)}= 2m_0$

Find Binding energy of an

$\alpha-$ particle in

$MeV$ ?

$[m_{proton}=1.007825\ amu, m_{neutron}=1.008665\ amu, m_{helium}=4.002800\ amu]$

0%
$28.097\ eV$
0%
$28.097\ MeV$
0%
$38.097\ eV$
0%
$48.097\ MeV$

Explanation

$\begin{array}{rl}\text { Given }- & * m_{\text {proton }}=1.007825\text { amu } \\&* m _{\text { neutron }}=1.008665 \text { amu } \\& * m_{\text {Helium }}=4.002800 \text { amu. }\end{array}$

$\text{For the reaction below-}$

$2 \cdot{ }_{1}^{1} \mathrm{H}+2 \cdot_{ 0}^{1} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}$

$\begin{array}{l}\text { We get ,Binding energy of }_{2}^{4}\mathrm{He} \\\Rightarrow B E=\Delta m \cdot c^{2}=\left[m_{\text {Helium }}-2 \mathrm{~m}_{\text {Proton }}-2\mathrm{~m}_{\text {neutron }}\right] c^{2}\\\qquad\begin{aligned}B E=&(0.03018) \text { amu } \times c^{2}=0.03018 \times 931\mathrm{MeV}\\&=28.097\mathrm{MeV}\end{aligned}\end{array}$

Calculate the binding energy of

$_{3}^{6}\textrm{Li}$ assuming the mass of

$_{3}^{6}\textrm{Li}$ atom as

$6.01512$ amu:

0%
$20.42 MeV$
0%
$30.42 MeV$
0%
$23.08 MeV$
0%
$32.78 MeV$

Explanation

Mass defect is

$\Delta m = [Zm_p + (A - Z)m_n-M]$

$m_p =$ mass of the proton

$= 1.007277 amu$

$m_n =$ mass of the neutron

$= 1.008655 amu$
Mass of the bound

$_{3}^{6}\textrm{Li}$ , lithium has three protons and three neutrons, since

$Z = 3, A = 6$ and number of neutrons

$=A -Z=6-3=3)$

$\therefore \Delta m = [3 \times 1.007277 + 3(1.008655] - 6.01512$

$=[3.02183 + 3.02596] - 6.01512$

$= 6.047795 - 6.01512 = 0.032675 amu.$
But the energy equivalent of

$1 amu = 931 MeV$
Therefore, the binding energy which is equal to the energy equivalent of mass defect in amu is

$E_b =\Delta m \times 931 MeV$

$= 0.032675 \times 931 = 30.42 MeV$

A neutron of kinetic energy

$65\ eV$ collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of

$90^{\circ}$ with respect to its original direction.
Find the allowed values of the energy of the neutron and that of the atom after the collision.
[Given : Mass of

$He\ atom = 4\times$ (mass of neutron) Ionization energy of

$H$ atom

$=6\ eV$ ]

0%
$7.36\ eV, 0.312, 17.8\ eV; 16.328\ eV$ .
0%
$6.36\ eV, 0.312, 17.8\ eV; 16.328\ eV$ .
0%
$6.36\ eV, 0.312, 87.8\ eV; 16.328\ eV$ .
0%
$6.36\ eV, 0.312, 17.8\ eV; 26.328\ eV$ .

Explanation

solution:

Applying conservation of linear momentum in horizontal direction

(Initial Momentum )x = (Final Momentum)x

$(P1)_x = (Pf)_x$

$\Rightarrow \sqrt{2}Km = \sqrt{2}(4m)K_1 cos θ …(i)$

Now applying conservation of linear momentum in Y – direction

$(Pi)_y = (Pf)_y$

$= \sqrt2K_2 m - \sqrt 2 (4m) K_1 sin θ$

$\rightarrow \sqrt 2 K_2 m = \sqrt 2(4m)K_1 sin θ …. (ii)$

Squaring and adding (i) and (ii)

$2Km + 2Km2 m = 2 (4m) K_1 + 2 (4m)K_1$

$K_1 + K_2 = 4K_1 ⇒ K = 4 K_1 – K_2 ⇒ 4K_1 – K_2 = 65 …(iii)$

When collision takes place, the electron gains energy and jumps to higher orbit.

Applying energy conservation

$K = K_1 + K_2 + \Delta E$

$65 = K_1 + K_2 + \Delta E$

Possible value of

$\Delta E$ for He+

Case (1)

$\Delta E_1 = - 13. 6 – (- 54.4 ) = 40.8 eV$

$K_1 + K_2 = 24.2 eV$ from (4)

Solving with (3), we get

$K_2 = 6.36 eV ; K_1 = 17 .84 eV$

Case (2)

$\Delta E_2 = - 6.04 – (- 54 .4) = 48 .36 eV$

$K_1 + K_2 = 16.64 eV$ from (4)

Solving with (3), we get $K_2 = 0.312 eV; K_1 = 16. 328 eV$

Hence the correct option: A

A system of binary stars of masses

$m_A$ and

$m_B$ are moving in circular orbits of radii

$r_A$ and

$r_B$ , respectively. If

$T_A$ and

$T_B$ are the time periods of masses

$m_A$ and

$m_B$ , respectively then.

0%
$\dfrac{T_A}{T_B}=\left(\dfrac{r_A}{r_B}\right)^{\dfrac{3}{2}}$
0%
$T_A > T_B$ (if $r_A > r_B$ )
0%
$T_A > T_B$ (if $m_A > m_B$ )
0%
$T_A = T_B$

Explanation

1) From the center of mass, we get

$M_{1}R_{1}=M_{2}R_{2}$ -------------1

2) Force acting on both stars i.e. gravitational force and centripetal force are balanced.

$\cfrac{GM_{1}M_{2}}{R^{2}}=\cfrac{M_{1}V_{1}^{2}}{R_{1}}=\cfrac{M_{1}V_{1}^{2}}{R_{1}}$ -------------2

As we know,

$V_{1}=\cfrac{2\pi R_{1}}{T_{1}}$ &

$V_{2}=\cfrac{2\pi R_{2}}{T_{2}}$

Substituting Equation 3 in Equation 2, we get

$\cfrac{M_{1}R_{1}}{(T_{1})^{2}}=\cfrac{M_{2}R_{2}}{(T_{2})^{2}}$ ------------4

From Equation 1, we conclude in Equation 4

$(T_{1})^{2}=(T_{2})^{2}$ or

$T_{1}=T_{2}$

${M}_{o}$ is the mass of an oxygen isotope

${ _{ 8 }^{ }{ O } }^{ 17 },{ M }_{ P },{ M }_{ N }$ are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is (The speed of light is C)

0%
$\left( { M }_{ O }-8{ M }_{ P } \right) { C }^{ 2 }$
0%
$\left( { M }_{ O }-8{ M }_{ P }-9{ M }_{ N } \right) { C }^{ 2 }$
0%
${ M }_{ O }{ C }^{ 2 }$
0%
$\left( { M }_{ O }-17{ M }_{ N } \right) { C }^{ 2 }$

Explanation

$\begin{aligned}\text { Given - }& *\text { Mass of } _8O^{17}\text { isotope }=M_{0} \\& *\text { Mass of neutron }=M_{N} \\& *\text { Mass of proton }=M_{p}\end{aligned}$

$\text{ For the decay of the Isotope we can write-}$

$_8O^{17} \longrightarrow{ }_{8}O^{16}+H^{+}(\text {Neutron) }$

$\text{ For the decay of the Isotope we can write-}$

$\begin{array}{l}B E=\Delta m \cdot c^{2}=\left(M_{\text {Reactant }}-M_{\text {product }}\right) \cdot C^{2} \\B E=\left(M_{0}-M_{\text {product }}\right) \cdot c^{2}\end{array}$

$\begin{array}{l}M \text { product } \Rightarrow \text { we have } 8\text { protons and } 9 \text { neutrons } \\M_{\text {product }}=8 M_{p}+9 M_{N} \\\Rightarrow \quad B E=\left(M_{0}-8 M_{p}-9 M_{N}\right) \cdot c^{2}\end{array}$

A nucleus

$^{A}_{Z}X$ has mass represented by

$m(A, Z)$ . If

$m_p$ and

$m_n$ denote the mass of proton and neutron respectively and BE the binding energy(in MeV) then.

0%
$BE=[m(A_1Z)-Zm_p-(A-Z)m_n]C^2$
0%
$BE=[Zm_p+(A-Z)m_n-m(A,Z)]C^2$
0%
$BE=[Zm_p+Am_n-m(A,Z)]C^2$
0%
$BE=m(A_1Z)-Zm_p-(A-Z)m_N$

Explanation

In the case of formation of nucleus the evolution of energy equals to the binding energy of the nucleus takes place due to disappearance of a fraction of total mass. If the quantity of mass disappearing is

$\Delta$ m, then the binding energy is

$BE=\Delta mc^2$
From the above discussion, it is clear that the mass of nucleus must be less than the sum of the masses of the constituent neutrons and protons. We can then write

$\Delta m=Zm_p-Nm_n-m(A, Z)$
where

$m(A, Z)$ is the mass of the atom of mass number A atomic number Z. Hence, the binding energy of nucleus is

$BE=[Zm_p+Nm_n-m(A, Z)]c^2$

$BE=[Zm_p+(A-Z)m_n-m(A, Z)]c^2$ .

Outside nucleus

0%
neutron is stable
0%
neutron is unstable
0%
proton and neutron both are stable
0%
none of these

Binding energies of

$_1H^2 , _2He^4 , _{26}Fe^{56} ,$ and

$_{92}U^{235}$ nuclie are

$2.22 Me V,28.4 MeV , 492MeV$ and

$1786MeV$ respectively which one of the following is more stable?

0%
$_1H^2$
0%
$_2He^4$
0%
$_{26}Fe^{56}$
0%
$_{92}U^{235}$

Explanation

The binding energy is the energy that is required to break the nucleons apart. The curve of Binding energy per nucleon Vs the Mass number shows the highest binding energy is that of Iron no other is tightly bound. Except the light nuclei having BE. ${\rm{56MeV}}$ .

Hence the most stable nucleon is ${\rm{2}}{\rm{.22MeV}}$

$\begin{matrix} M \\ Z \end{matrix}A(g)\longrightarrow \begin{matrix} M-B \\ Z-4 \end{matrix}B(g)+(\alpha -particals)$
(

$\alpha$ -particales are helium nuclei,so will form helium gas by trapping electrons)
The radioactive disintegration follows first-order kinetic Starting with 1 mol of A in a 1-litre closed flask at

$27^oC$ pressure developed after two half-lives is approximately:

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25 atm
0%
12 atm
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61.5 atm
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40atm

P and Q are two elements which form

${ P }_{ 2 }{ Q }_{ 3 }$ and

${ PQ }_{ 2 }$ . If 0.15 mole of

${ P }_{ 2 }{ Q }_{ 3 }$ weight 15.9 g and 0.15mole of

${ PQ }_{ 2 }$ weight 9.3 g. what are atomic weights of P and Q respectively?

0%
18 and 26
0%
26 and 26
0%
26 and 18
0%
18 and 18

Explanation

Using- no. of mole

$=\cfrac{\text {weight}}{\text{molecular weight}}$

$P_2Q_3 \Rightarrow$

$2P + 3Q =15.9/0.15= 106$ ....(1)

$PQ_2 \Rightarrow$

$P + 2Q = 9.3/0.15=62$ ........(2)

On solving eqn. (1) & (2) we get-

$P =26, Q = 18$

If the binding energy per nucleon in

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

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

$5.60\ MeV$ and

$7.06\ MeV$ respectively, then in the reaction :

$p + _{3}^{7}Li \rightarrow 2_{2}^{4}He$ energy of proton must be

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$28.24\ MeV$
0%
$17.28\ MeV$
0%
$1.46\ MeV$
0%
$39.2\ MeV$

Explanation

$\begin{array}{l}\text { Given }-B E \text { of }_{3}^{7} L_{i}=5.6\mathrm{MeV} , B E \text { of }_{2}^{4}\mathrm{He}=7.06\mathrm{MeV} \\p+\frac{7}{3}\mathrm{Li}\rightarrow 2 \cdot \frac{4}{2} \mathrm{He}\end{array}$

$\begin{aligned}\text { Energy of proton } &=2 \cdot B E \text { of }\mathrm{He}-\mathrm{BE} \text { of Li } \\&=2 \cdot[4 \times 7.06]-[7\times 5.60] \\&=17.28 \mathrm{MeV}\end{aligned}$

Assuming that

$200\ MeV$ of energy is released per fission of

$_{92}U^{235}$ atom. Find the number of fission per second ,required to release

$1\ kW$ power.

0%
$3.125\ \times 10^{13}$
0%
$3.125\ \times 10^{14}$
0%
$3.125\ \times 10^{15}$
0%
$3.125\ \times 10^{16}$

Explanation

$\begin{array}{l}\text { Given, } \\\text { Energy released per fission }=200 \times 10^{6}\mathrm{eV} \\\text { Power require }=1\times 10^{3} \text { watt } \\\text { Energy in Joule }=200 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J} \\\therefore 1\mathrm{ev}=1.6 \times 10^{-19} \mathrm{~J}\text { oule } \\\qquad=3.2 \times 10^{-11} \text {Joule }\end{array}$

$P=\frac{\text { Energy }}{time}$

$\text { let n : no of fission occur per second }$

$\text { Energy released will be }=n\left(3.2\times 10^{-11} \mathrm{~J}\right)$

$\begin{aligned}10^{3} &=n\left(3.2 \times 10^{-11}\right) \\n &=\frac{1}{3.2} \times 10^{14}\\&=0.3125 \times 10^{14} \\n &=3.125 \times 10^{13}\end{aligned}$

An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses ail its kinetic energy. Consequently the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The min. K.E. of colliding electron will be

0%
$10.2$ eV
0%
$1.9$ eV
0%
$12.1$ eV
0%
$13.6$ eV

Explanation

The electron hits the hydrogen atom in ground state with energy

$K$ .

This energy goes in exciting the atom. The electron loses all its energy.

Then, the atom emits a photon by jumping back to ground state.

So, the energy of the electron is equal to the energy of the photon emitted.

Now the maximum wavelength in the Baimer series is for the transition between

$3\to 2$ and so we get

$\dfrac {1}{\lambda_m} =R \left (\dfrac {1}{2^2} -\dfrac {1}{3^2}\right) =R \left (\dfrac {1}{4}-\dfrac {1}{9} \right)=\dfrac {5}{36}R$

Now, the energy of the photon emitted is

$E=\dfrac {hc}{\lambda_m}=\dfrac {hc5R}{36}=\dfrac {6.63 \times 10^{-34}\times 3\times 10^8 \times 5\times 1.097 \times 10^7}{36}$

$\therefore \ E=3.03\times 10^{-19}\ J$

$\therefore \ E=\dfrac {3.03\times 10^{-19}}{1.6\times 10^{-19}}=1.9\ eV$

Hence, the correct option is

$(B)$

In a laboratory experiment on emission from atomic hydrogen in a discharge tube, only a small number of lines are observed where as a lines are present in the hydrogen spectrum of a star. This is because in a laboratory

0%
The amount of hydrogen taken is much smaller than that present in the star
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The temperature of hydrogen is much smaller than that of the star
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The pressure of hydrogen is much smaller than that of the star
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The gravitational pull is much larger than that in the star

A nucleus

$_Z{X}^A$ emits

$9 \alpha$ -particles and

$5p$ particle. The ration of total protons and neutrons in the final nucleus is:-

0%
$\dfrac{(Z - 13)}{(A - Z - 23)}$
0%
$\dfrac{(Z - 18)}{(A - 36)}$
0%
$\dfrac{(Z - 23)}{(A - Z-8)}$
0%
$\dfrac{(Z - 13)}{(A - Z - 13)}$

Explanation

Hence, option

$C$ is correct answer.
1380174_1133378_ans_8ed5d27077a24b47988b7e13e983cf5e.jpg

$_{ a }^{ b }{ X }$ emits a positron, two

$\alpha$ and two

$\beta^-$ and in last one

$\alpha$ is also emitted and converts in

$_{ d }^{ c }{ Y }$ , correct relation:

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$c=b-12, d=a-5$
0%
$a=c-8, d=b-1$
0%
$a=c-6, d=b-0$
0%
$a=c-4, a=b-2$

$_{92}U^{238}$ on absorbing a neutron goes over to

$_{92}U^{239}$ . This nucleus emits an electron to go over to neptunium which on further emitting an electron goes over to plutonium. The plutonium nucleus can be expressed as:

0%
$_{94}Pu^{239}$
0%
$_{92}Pu^{239}$
0%
$_{93}Pu^{240}$
0%
$_{92}Pu^{240}$

Explanation

$\text{Given $-*_{92} U^{238}$ on absorbing neutron $\longrightarrow _{92} U^{239}$}$

$\text{Now the nucleus -}$

$_{92}{U^{239}} \quad \underline{\text { emits an } e^{-}} \quad _{93} N p^{239}$

$\begin{array}{l}\text { On further emitting an electron, } \\\qquad _{93}Np^{239} \quad \underline{\text { emits an } e^{-}} \quad _{94}Pu^{239} \\\text { Since, the emission of an } e^{-}\text{increases the atomic number } \\\text { by (1) and no change in the atomic mass. } \\\Rightarrow \text { opt (a) is correct. }\end{array}$

A radioactive nucleus

$_ZX^A$ emits

$3 \alpha$ -particles and

$5 \beta$ -particles. The ratio of number of neutron, protons in the product nucleus will be :-

0%
$\dfrac{A-Z-12}{Z-6}$
0%
$\dfrac{A-Z}{Z-1}$
0%
$\dfrac{A-Z-11}{Z-1}$
0%
$\dfrac{A-Z-12}{Z-1}$

Explanation

The new mass number after emission of

$3$ alpha particles will be

$A-3\times 4=A-12$

the new atomic number after emission of

$3$ alpha particles will be

$Z-3\times2=Z-6$

and after emission of

$5$ beta particles, this will become

$Z-6 +5\times 1=Z-1$

so the number of

$neutrons$ in the new nucleus will be

$A-12-(Z-1)=A-Z-11$

and the number of

$protons$ in the new nucleus will be

$(Z-1)$

the required ration will be

$\dfrac{n}{p}=\dfrac{A-Z-11}{Z-1}$

A nucleus of mass

$M$ is at rest. An alpha particle of mass

$m$ is emitted from the nucleus with momentum

$P. Q$ value of the nuclear reaction is :

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$\dfrac {p^{2}M}{2m(M + m)}$
0%
$\dfrac {p^{2}m}{2m(M + m)}$
0%
$\dfrac {p^{2}M}{2m(M - m)}$
0%
$\dfrac {p^{2}m}{2m(M - m)}$

Explanation

$\begin{array}{l} \text { Given, } \\ \text { nucleus of mass }=M \\ \text { mass of } \alpha \text { particle }=m \\ \text { momentem of } \alpha \text { -particle }=P \\ \therefore K E \text { of } \alpha \text { panticle }=\dfrac{p^{2}}{2 m} \end{array}$

$\begin{array}{l} \text { We know relation, } \\ \text { when } \alpha \text { paticle decay take place } \\ \qquad(K E)_{\alpha}=Q\left(\dfrac{A-\text { (mass of } \alpha )}{A}\right) \end{array}$

$Q=\dfrac{p^{2} M}{2 m(M-m)}$

$\begin{array} { l } { \text { Initial ratio of active nuclei in two different samples } } \\ { \text { is } 2 : 3 . \text { Their half lives are } 2 \text { hr and } 3 \text { hr respectively. } } \\ { \text { Ratio of their activities at the end of } 12 \text { hr is: } } \end{array}$

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$1 : 6$
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$6 : 1$
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$1 : 4$
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$4 : 1$

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

$E_{1}.$ Then the frequency of revolution of nth electron in the nth orbits is

0%
$\frac{2E_{1}}{nh}$
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$\frac{2E_{1}n^{3}}{h}$
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$\sqrt{\frac{2mE_{1}}{n^{3}h}}$
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$\frac{2E_{1}n^{2}}{h}$

Explanation

$\begin{array}{l}\text{B.E of electron= K.E of electron (in that orbit)}\\\dfrac{1}{2}mv^2=E_1-(i)\\\therefore \text{angular momentum -} mvr = \dfrac{nh}{2 \pi}-(ii)\\\text{Now from (i) divided by (ii)}\\\dfrac{v}{2r}= \dfrac{E_1 \times 2 \pi}{nh}\\\dfrac{v}{2 \pi r}=\dfrac{2E_1}{nh}\\\implies f=\dfrac{2E_1}{nh}\end{array}$

In the uranium radioactive series the initial nucleus is

$_{92} U^{238},$ and the final nucleus is

$_{82} U^{206}.$ When the Uranium nucleus decays to lead, the number of

$\alpha-particles$ emitted are... and the number of

$\beta-particles$ emitted are...

0%
$6, 8$
0%
$8, 6$
0%
$16, 6$
0%
$32, 12$

Explanation

$\begin{array}{l}\text { For each. } \alpha \text { -decay Mass number } \\\text { decreases by }{ }^{\prime\prime} 4^{\prime \prime} \text { and } Z \text { by " } 2^{\prime \prime} \\238 \longrightarrow 206\end{array}$

$\begin{array}{l}\text { Decrease in mass aumber is }{ }^{\prime\prime} 32^{\prime \prime} \\\text { Let number of }{ }^{\prime \prime} \alpha^{\prime \prime} \text { particles and }\\^{\prime\prime} \beta^{\prime \prime} \text { particles emitted be " } x^{\prime \prime} \text { and "y" } \\\qquad \begin{array}{c}4 x=32\\x=8\end{array}\end{array}$

$\begin{array}{l}\therefore \text { No. of } \alpha \text { -particles emitted will be } 8. \\\text { Now, due to this emission } z \text { decrease } \\\text { by } 16. \\\qquad \therefore \quad z=76 \\\quad 76\rightarrow 82 \\\therefore \quad y=6 \\\text { No of } \beta \text { -particles emitted is " } 6 ".\end{array}$

The aver age energy gy released in the fission of

$_{92}U^{235}$ is 200 MeV. The total energy released when one gram of

$_{92}U^{235}$ completely undergoes fission is about

0%
$2.3 \times 10^4 kWh$
0%
$2.3 \times 10^5 kWh$
0%
$2.3 \times 10^3 kWh$
0%
$2.3 \times 10^6 kWh$

In which sequence the radioactive radiations are emitted in the following nuclear reaction?

$_{Z}X^{A} \rightarrow _{Z+1}Y^{A} \rightarrow _{Z-1}K^{A-4} \rightarrow _{Z-1}K^{A-4}$

0%
$\gamma, \alpha ,\beta$
0%
$\alpha ,\beta,\gamma$
0%
$\beta,\gamma,\alpha$
0%
$\beta,\alpha,\gamma$

If the binding energy per nucleon in

$^7_3Li and ^4_2 He$ nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction

$p + ^7_3 Li \rightarrow 2^4_2 He$
energy of proton must be :

0%
39.2 MeV
0%
28.24 MeV
0%
17.28 MeV
0%
1.46 MeV

A nucleus of mass M +

$\triangle m$ is at rest and decays into daughter nucleus of equal mass

$\dfrac { M }{ 2 }$ each
speed of light is c
The speed of daughter nuclei is:-

0%
$c\sqrt { \dfrac { \triangle m }{ m+\triangle } }$
0%
$c\sqrt { \dfrac { \triangle m }{ m+\triangle m } }$
0%
$c\sqrt { \dfrac { 2\triangle m}{ m } }$
0%
$c\sqrt { \dfrac { \triangle m }{ m } }$

Atomic mass of

$_ { 26 } \mathrm { F } { \mathrm { e } } \text { is } 55.9349 \mathrm { u }$ and that of

$H \text { is } 1.00783 u$ .Mass of neutron is

$1.00867 \mathrm { u }$ and

$\mathrm { 1u } = 931 \mathrm { MeV } / \mathrm { c } ^ { 2 }$ then binding energy of

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

0%
$492 \mathrm { MeV }$
0%
$480 \mathrm { MeV }$
0%
$475 \mathrm { MeV }$
0%
$450 \mathrm { MeV }$

In the options given below, let E denote the rest mass energy of a nucleus and n neutron. The correct option is.

0%
$E\left( \begin{matrix} 236 \\ 92 \end{matrix}U \right) >E\left( \begin{matrix} 137 \\ 53 \end{matrix}I \right) +E\left( \begin{matrix} 97 \\ 39 \end{matrix}Y \right) +2E(n)$
0%
$E\left( \begin{matrix} 236 \\ 92 \end{matrix}U \right) \quad <\quad E\left( \begin{matrix} 137 \\ 53 \end{matrix}I \right) +E\left( \begin{matrix} 97 \\ 39 \end{matrix}Y \right) +2E(n)$
0%
$E\left( \begin{matrix} 236 \\ 92 \end{matrix}U \right) <\quad E\left( \begin{matrix} 140 \\ 56 \end{matrix}Ba \right) +E\left( \begin{matrix} 94 \\ 36 \end{matrix}Kr \right) +2E(n)$
0%
$E\left( \begin{matrix} 236 \\ 92 \end{matrix}U \right) =\quad E\left( \begin{matrix} 140 \\ 56 \end{matrix}Ba \right) +E\left( \begin{matrix} 94 \\ 36 \end{matrix}Kr \right) +2E(n)$

In the nuclear reaction,

$X\left( {n,\alpha } \right){ \to _3}L{i^7}$ which of the following option is correct?

0%
$X{ = _5}{\beta ^{10}}$ , $\alpha$ -deacy result in decrease of atomic number
0%
$X{ = _5}{\beta ^{10}}$ , only ${\beta ^ + }$ - decay result in increase of atomic number
0%
$X{ = _5}{\beta ^{11}}$ , $\alpha$ -decay result in increase of atomic number
0%
$X{ = _5}{\beta ^{11}}$ , $\alpha$ - and ${\beta ^ + }$ -decay result in decrease of atomic number

The binding energies per nucleon of deutron and

$\alpha - particles$ are

$X_1$ and

$X_2$ respectively. The energy released in the following reaction will be:

$1{H}^2 + 1{H}^2 = 2{He}^4 + Q$

0%
$(X_1 + X_2)$
0%
$(X_2 - X_1)$
0%
$4(X_1 + X_2)$
0%
$4(X_2 - X_1)$

Binding energy per nucleon of deutron in 1.112 MeV and binding energy per nucleon of a

$\alpha$ -particle is 7.07 MeV, then in following process, energy Q is :-

$2(1_H^2)\rightarrow _2He^4+ Q$

0%
1 MeV
0%
23.8 MeV
0%
11.9 MeV
0%
931 MeV

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

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Practice Class 12 Medical Physics Quiz Questions and Answers

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

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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