Q: \$M, M\$\$_{n}\$ and \$M\$\$_{p}\$ denotes the masses of a nucleus of \$_{Z}X^{A}\$, a neutron, and a proton respectively. If the nucleus is separated into its individual protons and neutrons then,

\$ Mc^{2} + Binding energy = [( A-Z)Mn + Z Mp] c^{2}\$Therefore, mass of nucleus is less than total mass of its free nucleons.

Q: The binding energies of the atoms of elements \$P\$ and \$Q\$ are \$E_{p}\$ and \$E_{Q}\$, respectively. Three atoms of element \$Q\$ fuse to form one atom of element \$P\$. In this process, the energy released is \$e\$. The correct relation between \$E_{P}, E_{Q}\$ and \$e\$ will be

Binding energy of \$3\$ \$Q\$ elements will be \$3E_Q\$So, by energy conservation, binding energy of three \$P\$ elements is equal to the sum of \$Q\$ element's binding energy and the released energy.So, equation is \$E_p + e = 3E_Q \$\$E_p = 3E_Q - e \$

Q: In nuclear reaction \$_{4}Be^{9}+_{2}He^{4} \rightarrow _{6}c^{12}\$+X, X will be :

\$_{4}Be^{9} + _{2}He^{4} \rightarrow _{6}C^{12} + X\$No. of protons are already balanced with \$C\$.\$\therefore\$ \$X\$ carries no charge.and \$ 9+4 > 12\$ by \$1\$ unit mass.\$\therefore\$ \$X\$ carries one unit mass.It's a neutron.

Q: The particle A is converted to C via following reactions then :\$A \rightarrow B + _{2}He^{4}\$\$B \rightarrow C + 2 _{-1}e^{0}\$

No of neutrons \$ = N_A N_B N_C\$No of protons \$ = Z_A Z_B Z_C\$Mass \$ = N_A+Z_A N_B+Z_B N_C+Z_C\$ For reaction \$ A \rightarrow B + _{2}He^{4}\$ \$ Z_A = Z_B + 2 ............. [1] \$For reaction \$B \rightarrow C + 2 _{-1}e^{0}\$ \$ Z_B = Z_C - 2 ..............[2]\$ Solving these two equations, we get: \$ Z_A = Z_C\$ Hence they are isotopes.

Q: On the bombardment of Boron with neutron, an \$\alpha \$ - particle is emitted and product nucleus formed is \$\underline{\hspace{0.5in}}\$

\$ _{5}B^{10}+^{1}n \rightarrow _{Z}^{A}X + _{2}^{4}He\$Mass balance: \$ 10 + 1 = A + 4\$Atomic no balance: \$ 5 + 0 = Z + 2\$ \$ Z = 3\$Therefore, it's \$ _{3}Li^{7}\$

Q: The missing particle in the reaction : \$^{253}_{99}Es+^{4}_{2}He\rightarrow ^{256}_{101}Md+ \underline{ \hspace{0.5in}}\$

Let the particle be \$X\$.On summing the mass number on left hand side we get \$253+4 = 257\$ units.On right hand side we have \$256\$ units only. So, we should have \$1\$ unit mass number on \$X\$ element.On summing the atomic number on left hand side we get \$99 + 2 = 101\$ unitsOn right hand side we have \$101\$ units of atomic number. So, we should have \$0\$ unit atomic number on \$X\$ element.So \$^1_0X = ^1_0n\$ i.e. neutron.

Question 1

A proton and a neutron combine to give a deuterium nucleus.If $m_{o}$ and  $m_{p}$ be the mass of neutron and proton respectively, then mass of deuterium nucleus is

Accepted Answer

The energy released during this during this in form of gamma photon comes from mass defect. (i.e., $E=mc^2$, where $m$ is the mass defect). The mass of the deuterium nucleus (2.01355 u) is less than the sum of the masses of the proton (1.00728 u) and the neutron (1.00866 u), which is 2.01594 u.

Question 2

In a nuclear reaction some mass converts into energy. In this reaction total B.E of reactants when compared with that of products is:

Accepted Answer

The difference in mass between the separate particles and the nuclide is called the mass defect. This 'missing mass' accounts for the binding energy..Binding&#10; energy of the products is always greater than the binding energy of the&#10; reactants and the difference is released as energy.

Question 3

$M, M$$_{n}$ and $M$$_{p}$ denotes the masses of a nucleus of $_{Z}X^{A}$, a neutron, and a proton respectively. If the nucleus is separated into its individual protons and neutrons then,

Accepted Answer

$ Mc^{2} +  Binding   energy = [( A-Z)Mn + Z Mp] c^{2}$Therefore, mass of nucleus is less than total mass of its free nucleons.

Question 4

The nucleus finally formed in fusion of the proton in a proton cycle is that of:

Accepted Answer

Proton - Proton cycle :The thermonuclear reactions involved are :$2(_1^1H) + 2(_1^1H) \Rightarrow 2(_1^2H) + 2(_1^0e) + Q_1$$2(_1^2H) + 2(_1^1H) \Rightarrow 2(_2^3He) + Q_2$$2(_2^3He)  \Rightarrow (_2^4He) + 2(_1^2H) + Q_3$On adding up these reactions, we obtain$4(_1^1H)  \Rightarrow _2^4He + 2(_1^0e) + Q$where, $Q = Q_1 + Q_2 + Q_3$ is the total energy involved in the fusion of 4 hydrogen nuclei to form Helium nucleus. The value of Q as calculated from mass defect comes out to be 26.7 MeV.

Question 5

The binding energies of the atoms of elements $P$ and $Q$ are $E_{p}$ and $E_{Q}$ , respectively. Three atoms of element $Q$ fuse to form one atom of element $P$ . In this process, the energy released is $e$ . The correct relation between $E_{P}, E_{Q}$ and $e$ will be

Accepted Answer

Binding energy of $3$ $Q$ elements will be $3E_Q$So, by energy conservation, binding energy of three $P$ elements is equal to the sum of $Q$ element's binding energy and the released energy.So, equation is $E_p + e  = 3E_Q $$E_p  = 3E_Q - e $

Question 6

In nuclear reaction $_{4}Be^{9}+_{2}He^{4} \rightarrow _{6}c^{12}$+X,  X will be :

Accepted Answer

$_{4}Be^{9} + _{2}He^{4} \rightarrow  _{6}C^{12} + X$No. of protons are already balanced with $C$.$\therefore$ $X$ carries no charge.and $ 9+4 > 12$ by $1$ unit mass.$\therefore$ $X$ carries one unit mass.It's a neutron.

Question 7

Among the following reactions , the impossible one is :

Accepted Answer

For possible reactions :Sum of mass numbers LHS should be equal to the sum of mass numbers on RHS of the equation.As well as sum of Atomic numbers on LHS should be equal to the sum of Atomic numbers on RHS of the equation.So, in option C we can see that on LHS mass number is 4 x 1 = 4 and Atomic number is 4 x 1 = 4. on RHS mass number is 2 + 2(-1) = 0 and Atomic number is 4 + 0 = 4.So, mass number are different in LHS and RHS of the equation. So, this reaction is not possible.

Question 8

The particle A is converted to C via following reactions then :$A \rightarrow  B +  _{2}He^{4}$$B \rightarrow  C + 2  _{-1}e^{0}$

Accepted Answer

No of neutrons $ =                      N_A             N_B               N_C$No of protons  $  =                      Z_A             Z_B                Z_C$Mass $ =                   N_A+Z_A       N_B+Z_B        N_C+Z_C$  For reaction  $ A  \rightarrow   B + _{2}He^{4}$                     $ Z_A = Z_B + 2    ............. [1] $For reaction  $B  \rightarrow  C + 2 _{-1}e^{0}$                   $  Z_B = Z_C - 2      ..............[2]$             Solving these two equations, we get:                     $ Z_A = Z_C$                       Hence they are isotopes.

Question 9

On the bombardment of Boron with neutron, an $\alpha $ - particle is emitted and product nucleus formed is $\underline{\hspace{0.5in}}$

Accepted Answer

$ _{5}B^{10}+^{1}n \rightarrow  _{Z}^{A}X + _{2}^{4}He$Mass balance: $ 10 + 1  = A + 4$Atomic no balance: $ 5 + 0 = Z + 2$                                 $ Z = 3$Therefore,  it's $ _{3}Li^{7}$

Question 10

The missing particle in the reaction : $^{253}_{99}Es+^{4}_{2}He\rightarrow  ^{256}_{101}Md+ \underline{ \hspace{0.5in}}$

Accepted Answer

Let the particle be $X$.On summing the mass number on left hand side we get $253+4 = 257$ units.On right hand side we have $256$ units only. So, we should have $1$ unit mass number on $X$ element.On summing the atomic number on left hand side we get $99 + 2 = 101$ unitsOn right hand side we have $101$ units of atomic number. So, we should have $0$ unit atomic number on $X$ element.So $^1_0X =  ^1_0n$ i.e. neutron.

Nuclei

Physics
4585 Views

A proton and a neutron combine to give a deuterium nucleus.If $m_{o}$ and $m_{p}$ be the mass of neutron and proton respectively, then mass of deuterium nucleus is

In a nuclear reaction some mass converts into energy. In this reaction total B.E of reactants when compared with that of products is:

$M, M$ $_{n}$ and $M$ $_{p}$ denotes the masses of a nucleus of $_{Z} X^{A}$ , a neutron, and a proton respectively. If the nucleus is separated into its individual protons and neutrons then,

The nucleus finally formed in fusion of the proton in a proton cycle is that of:

The binding energies of the atoms of elements $P$ and $Q$ are $E_{p}$ and $E_{Q}$ , respectively. Three atoms of element $Q$ fuse to form one atom of element $P$ . In this process, the energy released is $e$ . The correct relation between $E_{P}, E_{Q}$ and $e$ will be

In nuclear reaction $_{4} B e^{9} +_{2} H e^{4} \to_{6} c^{12}$ +X, X will be :

Among the following reactions , the impossible one is :

The particle A is converted to C via following reactions then :
$A \to B +_{2} H e^{4}$
$B \to C + 2_{- 1} e^{0}$

On the bombardment of Boron with neutron, an $α$ - particle is emitted and product nucleus formed is

The missing particle in the reaction :
$_{99}^{253} E s +_{2}^{4} H e \to_{101}^{256} M d +$

Nuclei

Physics 4585 Views

A proton and a neutron combine to give a deuterium nucleus.If mo​ and mp​ be the mass of neutron and proton respectively, then mass of deuterium nucleus is

In a nuclear reaction some mass converts into energy. In this reaction total B.E of reactants when compared with that of products is:

M,Mn​ and Mp​ denotes the masses of a nucleus of Z​XA, a neutron, and a proton respectively. If the nucleus is separated into its individual protons and neutrons then,

The nucleus finally formed in fusion of the proton in a proton cycle is that of:

The binding energies of the atoms of elements P and Q are Ep​ and EQ​, respectively. Three atoms of element Q fuse to form one atom of element P. In this process, the energy released is e. The correct relation between EP​,EQ​ and e will be

In nuclear reaction 4​Be9+2​He4→6​c12+X, X will be :

Among the following reactions , the impossible one is :

The particle A is converted to C via following reactions then :A→B+2​He4B→C+2−1​e0

On the bombardment of Boron with neutron, an α - particle is emitted and product nucleus formed is ​

The missing particle in the reaction : 99253​Es+24​He→101256​Md+​

Physics
4585 Views

A proton and a neutron combine to give a deuterium nucleus.If $m_{o}$ and $m_{p}$ be the mass of neutron and proton respectively, then mass of deuterium nucleus is

$M, M$ $_{n}$ and $M$ $_{p}$ denotes the masses of a nucleus of $_{Z} X^{A}$ , a neutron, and a proton respectively. If the nucleus is separated into its individual protons and neutrons then,

The binding energies of the atoms of elements $P$ and $Q$ are $E_{p}$ and $E_{Q}$ , respectively. Three atoms of element $Q$ fuse to form one atom of element $P$ . In this process, the energy released is $e$ . The correct relation between $E_{P}, E_{Q}$ and $e$ will be

In nuclear reaction $_{4} B e^{9} +_{2} H e^{4} \to_{6} c^{12}$ +X, X will be :

The particle A is converted to C via following reactions then :
$A \to B +_{2} H e^{4}$
$B \to C + 2_{- 1} e^{0}$

On the bombardment of Boron with neutron, an $α$ - particle is emitted and product nucleus formed is

The missing particle in the reaction :
$_{99}^{253} E s +_{2}^{4} H e \to_{101}^{256} M d +$