Q: In the reaction \$_7N^{14} +\$ \$ _2He^4 \rightarrow _8O^{17} + \$ \$_1H^1\$. The minimum energy of \$\alpha\$-particle is\$M_N=\$ \$14.00307\$ amu \$M_{He}=\$ \$4.00260\$ amu\$M_o =\$ \$16.99914\$ amu \$M_H=\$ \$1.00783\$ amu

Mass on reactant side \$= (14.00307 + 4.00260) amu \$Mass on product side \$ = (16.99914 + 1.00783) amu \$So, mass defect \$ = 1.3 \times10^{-3} amu.\$Energy of the \$\alpha \$ particle \$= 931.5\times1.3\times10^{-3} MeV\$ \$ = 1.21 MeV\$

Q: The energy required to separate the typical middle mass nucleus \$^{120}_{50}Sn\$ into its constituent nucleons is :(Mass of \$^{120}_{50}Sn=\$ \$119.902199\ amu\$; massof proton\$=1.007825\ amu\$ and mass of neutron\$=1.008665\ amu\$)

\$Z = 50, A-Z = 120-50 = 70\$\$\Delta m = Z.m_p + (A-Z)m_n - M\$\$\Delta m = [50\times 1.007825 + 70\times 1.008665 - 119.902199]\$ \$= 1.095601\ MeV\$\$E = 1.095601\times 931.478\ MeV\$ \$=1020.53\ MeV\$ \$\approx\$ \$1021\ MeV\$

Q: The kinetic energy (in keV) of the alpha particle, when the nucleus \$^{210}_{84}Po\$ at rest undergoes alpha decay, is

\$^{210}_{84}Po\rightarrow_{2}^{4}\mathrm{H}\mathrm{e}+_{82}^{206}\$ Pb\$\mathrm{Q}=(209.982876-4.002603-205.97455)\mathrm{C}^{2}\$ \$=5.422\mathrm{M}\mathrm{e}\mathrm{V}\$From conservation of momentum:\$\sqrt{2\mathrm{K}_{1}(4)}=\sqrt{2\mathrm{K}_{2}(206)}\$\$4\mathrm{K}_{1}=206\mathrm{K}_{2}\$\$\displaystyle \mathrm{K}_{1}=\frac{103}{2}\mathrm{K}_{2}\$\$\mathrm{K}_{1}+\mathrm{K}_{2}=5.422\$\$\displaystyle \mathrm{K}_{1}+\frac{2}{103}\mathrm{K}_{1}=5.422\$\$\Rightarrow \displaystyle \frac{105}{103}\mathrm{K}_{1}=5.422\$\$\mathrm{K}_{1}=5.319\mathrm{M}\mathrm{e}\mathrm{V}=5319\mathrm{K}\mathrm{e}\mathrm{V}\$

Q: Atomic masses of \$^{14}_{7}N\$ and \$^{16}_{8}O\$ are \$14.008\$ amu and \$16.000\$ amu respectively. The mass of \$^{1}_{1}H\$ atom is 1.007825 amu and the mass of neutron is \$1.008665\$ amu. Pick the correct option

Mass of 7 protons and 7 neutrons=\$7(1.007825+1.008665)=14.11543amu\$Mass of \$_7^{14}N=14.008amu\$Hence mass lost as energy in formation of nitrogen atom=\$0.10743amu\$Mass of 8 protons and 8 neutrons=\$8(1.007825+1.008665)amu=16.13192amu\$Mass of \$_8^{16}O=16.000amu\$Hence mass lost as energy in formation of oxygen atom=\$0.13192amu\$Since more mass has been lost in formation of oxygen atom, oxygen atom is more stable.

Q: The mass of chlorine (\$_{17}Cl^{35}\$) atom is 34.98 amu, mass of proton = 1.007825 amu, mass of neutron= 1.008665 amu. Then binding energy is :

\$ \Delta m = (17 \times 1.007825) + (18 \times 1.008665) - 34.98\$ \$ = 0.308995 amu\$ \$ = 0.308995 \times 931.478 \dfrac{MeV}{c^{2}} ....... [\$using \$1 amu = 931.478 \dfrac{MeV}{c^{2}}]\$ \$ = 287.82 \dfrac{MeV}{c^{2}}\$\$ E = \Delta m c^{2}\$ \$ = 287.82 MeV\$

Q: The \$\beta\$-decay process, discovered around 1900, is basically the decay of a neutron \$(n)\$. In the laboratory, a proton \$(p)\$ and an electron \$(e^{-})\$ are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e. \$n\rightarrow p+e^{-}+\overline{v}_{e}\$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino \$(\overline{v}_{e})\$ to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is \$0.8\times 10^{6}eV\$. The kinetic energy carried by the proton is only the recoil energy.If the anti-neutrino had a mass of 3 \$eV/c^{2}\$ (where \$c\$ is the speed of light) instead of zero mass, what should be the range of the kinetic energy, \$K\$, of the electron?

\$K\$ can't be equal to \$0.8 \times 10^6 eV\$ as anti-neutrino must have some energy

Q: Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy \$1.5 kT\$, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature \$T\$ required for them to reach a separation of \$4 \times 10^{-15}\$ m is in the range

Conservation of energy:\$ KE = PE\$\$ 2 (1.5kT) = \cfrac{e^2}{4 \pi \epsilon _o d}\$\$ 3kT = \cfrac{(1.6 \times 10^{-19})^2}{4 \pi (8.84 \times 10^{-12} (4 \times 10^{-15}}\$\$ T =1.4 \times 10^9 K\$

Q: A radioactive nucleus undergoes a series of decays according to the sequence\$A\overset{\beta}{\rightarrow}A_1\overset{\alpha}{\rightarrow}A_2\overset{\alpha}{\rightarrow}A_3\$If the mass number and atomic number of \$A_3\$ are \$172\$ and \$69 \$respectively, then the mass number and atomic number of A is

\$^A_Z{A} \rightarrow A_1 \rightarrow A_2 \rightarrow _69 {A}_3^{172}\$Mass balance, \$A = 0 + 4 + 4 + 172\$ \$= 180\$Charge balance, \$Z= -1 + 2 + 2 + 69\$ \$ = 72\$

Question 1

The correct statement is

Accepted Answer

For fusion reaction to take place, sum of masses of lighter nuclei must be greater than mass of heavy nucleus formed.For fusion reaction :     $^2_1H + ^4_2He\rightarrow ^6_3Li$Sum of mass of lighter nuclei     $M_l = 2.014102 + 4.002603  = 6.016705$ uMass of  $^6_3Li$     $M_h = 6.015123$ uAs  $M_l>M_h$, thus above reaction is possible.

Question 2

In the reaction $_7N^{14} +$ $ _2He^4 \rightarrow  _8O^{17} + $ $_1H^1$. The minimum energy of $\alpha$-particle is$M_N=$ $14.00307$ amu      $M_{He}=$ $4.00260$ amu$M_o =$ $16.99914$ amu      $M_H=$ $1.00783$ amu

Accepted Answer

Mass on reactant side $= (14.00307 + 4.00260) amu $Mass on product side $ = (16.99914 + 1.00783) amu $So, mass defect $ = 1.3 \times10^{-3}  amu.$Energy of the $\alpha $ particle $= 931.5\times1.3\times10^{-3} MeV$                                         $ = 1.21 MeV$

Question 3

The energy required to separate the typical middle mass nucleus $_{50}^{120} S n$ into its constituent nucleons is :
(Mass of $_{50}^{120} S n =$ $119.902199 a m u$ ; mass
of proton $= 1.007825 a m u$ and mass of neutron $= 1.008665 a m u$ )

Accepted Answer

$Z = 50, A-Z = 120-50 = 70$$\Delta m = Z.m_p + (A-Z)m_n - M$$\Delta m = [50\times 1.007825 + 70\times 1.008665 - 119.902199]$        $= 1.095601\  MeV$$E = 1.095601\times 931.478\ MeV$    $=1020.53\  MeV$ $\approx$ $1021\ MeV$

Question 4

The kinetic energy (in keV) of the alpha particle, when the nucleus $^{210}_{84}Po$ at rest undergoes alpha decay, is

Accepted Answer

$^{210}_{84}Po\rightarrow_{2}^{4}\mathrm{H}\mathrm{e}+_{82}^{206}$ Pb$\mathrm{Q}=(209.982876-4.002603-205.97455)\mathrm{C}^{2}$    $=5.422\mathrm{M}\mathrm{e}\mathrm{V}$From conservation of momentum:$\sqrt{2\mathrm{K}_{1}(4)}=\sqrt{2\mathrm{K}_{2}(206)}$$4\mathrm{K}_{1}=206\mathrm{K}_{2}$$\displaystyle \mathrm{K}_{1}=\frac{103}{2}\mathrm{K}_{2}$$\mathrm{K}_{1}+\mathrm{K}_{2}=5.422$$\displaystyle \mathrm{K}_{1}+\frac{2}{103}\mathrm{K}_{1}=5.422$$\Rightarrow  \displaystyle \frac{105}{103}\mathrm{K}_{1}=5.422$$\mathrm{K}_{1}=5.319\mathrm{M}\mathrm{e}\mathrm{V}=5319\mathrm{K}\mathrm{e}\mathrm{V}$

Question 5

Atomic masses of $^{14}_{7}N$ and $^{16}_{8}O$ are $14.008$ amu and $16.000$ amu respectively. The mass of $^{1}_{1}H$ atom is 1.007825 amu and the mass of neutron is $1.008665$ amu. Pick the correct option

Accepted Answer

Mass of 7 protons and 7 neutrons=$7(1.007825+1.008665)=14.11543amu$Mass of $_7^{14}N=14.008amu$Hence mass lost as energy in formation of nitrogen atom=$0.10743amu$Mass of 8 protons and 8 neutrons=$8(1.007825+1.008665)amu=16.13192amu$Mass of $_8^{16}O=16.000amu$Hence mass lost as energy in formation of oxygen atom=$0.13192amu$Since more mass has been lost in formation of oxygen atom, oxygen atom is more stable.

Question 6

The mass of chlorine ($_{17}Cl^{35}$) atom is 34.98 amu, mass of proton = 1.007825 amu, mass of neutron= 1.008665 amu. Then binding energy is :

Accepted Answer

$ \Delta m = (17 \times 1.007825) + (18 \times 1.008665) - 34.98$        $ = 0.308995  amu$        $ = 0.308995 \times 931.478  \dfrac{MeV}{c^{2}} .......         [$using $1   amu    =  931.478 \dfrac{MeV}{c^{2}}]$        $ = 287.82  \dfrac{MeV}{c^{2}}$$ E = \Delta m c^{2}$    $ = 287.82  MeV$

Question 7

The $\beta$-decay process, discovered around 1900, is basically the decay of a neutron $(n)$. In the laboratory, a proton $(p)$ and an electron $(e^{-})$ are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e. $n\rightarrow p+e^{-}+\overline{v}_{e}$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $(\overline{v}_{e})$ to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is $0.8\times 10^{6}eV$. The kinetic energy carried by the proton is only the recoil energy.If the anti-neutrino had a mass of 3 $eV/c^{2}$ (where $c$ is the speed of light) instead of zero mass, what should be the range of the kinetic energy, $K$, of the electron?

Accepted Answer

$K$ can't be equal to $0.8 \times 10^6 eV$ as anti-neutrino must have some energy

Question 8

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

Accepted Answer

The gas becomes plasma because of the high temperature. The kinetic energy of the particles are directly proportional to absolute temperature, the high temperature causes the particles to turn into plasma (from gas).Hence option D is correct.

Question 9

Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy $1.5 kT$, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature $T$ required for them to reach a separation of $4 \times 10^{-15}$ m is in the range

Accepted Answer

Conservation of energy:$ KE  = PE$$ 2 (1.5kT) = \cfrac{e^2}{4 \pi \epsilon _o d}$$ 3kT =  \cfrac{(1.6 \times 10^{-19})^2}{4 \pi (8.84 \times 10^{-12} (4 \times 10^{-15}}$$ T =1.4 \times 10^9 K$

Question 10

A radioactive nucleus undergoes a series of decays according to the sequence
$A \to β A_{1} \to α A_{2} \to α A_{3}$
If the mass number and atomic number of $A_{3}$ are $172$ and $69$ respectively, then the mass number and atomic number of A is

A

56, 23

B

180, 72

C

120, 52

D

84, 38

Hard

View solution>

Accepted Answer

$^A_Z{A} \rightarrow A_1 \rightarrow A_2 \rightarrow _69 {A}_3^{172}$Mass balance, $A = 0 + 4 + 4 + 172$                             $= 180$Charge balance, $Z= -1 + 2 + 2 + 69$                                 $ = 72$

Nuclei

Physics
3361 Views

The correct statement is

In the reaction $_{7} N^{14} +$ $_{2} H e^{4} \to_{8} O^{17} +$ $_{1} H^{1}$ . The minimum energy of $α$ -particle is
$M_{N} =$ $14.00307$ amu $M_{H e} =$ $4.00260$ amu
$M_{o} =$ $16.99914$ amu $M_{H} =$ $1.00783$ amu

The energy required to separate the typical middle mass nucleus $_{50}^{120} S n$ into its constituent nucleons is :
(Mass of $_{50}^{120} S n =$ $119.902199 a m u$ ; mass
of proton $= 1.007825 a m u$ and mass of neutron $= 1.008665 a m u$ )

The kinetic energy (in keV) of the alpha particle, when the nucleus $_{84}^{210} P o$ at rest undergoes alpha decay, is

Atomic masses of $_{7}^{14} N$ and $_{8}^{16} O$ are $14.008$ amu and $16.000$ amu respectively. The mass of $_{1}^{1} H$ atom is 1.007825 amu and the mass of neutron is $1.008665$ amu. Pick the correct option

The mass of chlorine ( $_{17} C l^{35}$ ) atom is 34.98 amu, mass of proton = 1.007825 amu, mass of neutron= 1.008665 amu. Then binding energy is :

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

A radioactive nucleus undergoes a series of decays according to the sequence
$A \to β A_{1} \to α A_{2} \to α A_{3}$
If the mass number and atomic number of $A_{3}$ are $172$ and $69$ respectively, then the mass number and atomic number of A is

Nuclei

Physics 3361 Views

The correct statement is

In the reaction 7​N14+ 2​He4→8​O17+ 1​H1. The minimum energy of α-particle is MN​= 14.00307 amu MHe​= 4.00260 amu Mo​= 16.99914 amu MH​= 1.00783 amu

The energy required to separate the typical middle mass nucleus 50120​Sn into its constituent nucleons is : (Mass of 50120​Sn= 119.902199 amu; mass of proton=1.007825 amu and mass of neutron=1.008665 amu)

The kinetic energy (in keV) of the alpha particle, when the nucleus 84210​Po at rest undergoes alpha decay, is

Atomic masses of 714​N and 816​O are 14.008 amu and 16.000 amu respectively. The mass of 11​H atom is 1.007825 amu and the mass of neutron is 1.008665 amu. Pick the correct option

The mass of chlorine (17​Cl35) atom is 34.98 amu, mass of proton = 1.007825 amu, mass of neutron= 1.008665 amu. Then binding energy is :

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

A radioactive nucleus undergoes a series of decays according to the sequence A→βA1​→αA2​→αA3​ If the mass number and atomic number of A3​ are 172 and 69respectively, then the mass number and atomic number of A is

Physics
3361 Views

In the reaction $_{7} N^{14} +$ $_{2} H e^{4} \to_{8} O^{17} +$ $_{1} H^{1}$ . The minimum energy of $α$ -particle is
$M_{N} =$ $14.00307$ amu $M_{H e} =$ $4.00260$ amu
$M_{o} =$ $16.99914$ amu $M_{H} =$ $1.00783$ amu

The energy required to separate the typical middle mass nucleus $_{50}^{120} S n$ into its constituent nucleons is :
(Mass of $_{50}^{120} S n =$ $119.902199 a m u$ ; mass
of proton $= 1.007825 a m u$ and mass of neutron $= 1.008665 a m u$ )

The kinetic energy (in keV) of the alpha particle, when the nucleus $_{84}^{210} P o$ at rest undergoes alpha decay, is

Atomic masses of $_{7}^{14} N$ and $_{8}^{16} O$ are $14.008$ amu and $16.000$ amu respectively. The mass of $_{1}^{1} H$ atom is 1.007825 amu and the mass of neutron is $1.008665$ amu. Pick the correct option

The mass of chlorine ( $_{17} C l^{35}$ ) atom is 34.98 amu, mass of proton = 1.007825 amu, mass of neutron= 1.008665 amu. Then binding energy is :

A radioactive nucleus undergoes a series of decays according to the sequence
$A \to β A_{1} \to α A_{2} \to α A_{3}$
If the mass number and atomic number of $A_{3}$ are $172$ and $69$ respectively, then the mass number and atomic number of A is