Tag: nuclear reactions

Questions Related to nuclear reactions

The binding energy per nucleon of deutron $(^2 _1 H)$ and helium nucleus $(^4 _2 He)$ is 1.1 MeV and 7 MeV respectively. If two deutron nuclei react to form a single helium nucleus, then the energy released is

  1. $23.6 MeV$

  2. $26.9 MeV$

  3. $13.9 MeV$

  4. $19.2 MeV$


Correct Option: A

Binding energy per nucleon is $8.5 \text { MeV for } A = 120$ and is $7.6 \mathrm { MeV } \text { for } \mathrm { A } = 240$ Suppose a nucleus with $A = 240$ breaks into two nuclei of nearly equal mass numbers then which of the following is correct

  1. 216 MeV energy is released.

  2. 21 MeV energy is to be given from outside

  3. 220 MeV energy is released.

  4. no energy is released.


Correct Option: A

Energy released if mass of $2\ amu$ is converted into energy is :

  1. $1.5 \times 10^{-10}\ J$

  2. $3 \times 10^{-10}\ J$

  3. $1863\ J$

  4. $931.5 \Mev$


Correct Option: B
Explanation:

$ E = \Delta m c^{2}$
    $ = (2 \times 1.67 \times 10^{-27}  kg) \times (3 \times 10^{8} \frac{m}{s})^{2} $
    $ = 3 \times 10^{-10}  J$

When an electron and a positron are annihilated, then the number of photons produced is

  1. 2

  2. 1

  3. 3

  4. 4


Correct Option: A
Explanation:

Two photons are produced during the annihilation of an electron and a positron along with $1.02$ MeV released energy.

Consider the nuclear reaction: $\mathrm { X } ^ { 200 } \longrightarrow \mathrm { A } ^ { 110 } + \mathrm { B } ^ { 20 }$If the binding energy per nucleon for $\mathrm { X } , \mathrm { A }$ and $\mathrm { B }$ is $7.4 \mathrm { MeV } , 8.2 \mathrm { MeV }$ and 8.2$\mathrm { MeV }$ respectively, what is the energy relesed?

  1. $200$ $\mathrm { MeV }$

  2. $160$ $\mathrm { MeV }$

  3. $110$ $\mathrm { MeV }$

  4. $90$  $\mathrm { MeV }$


Correct Option: B

In the nucleus of helium if ${ F } _{ 1 }$ is the net force between two protons ${ F } _{ 2}$ is the net force between two neutrons and ${ F } _{ 3 }$ is the net force between a proton and a neutron. Then,

  1. ${ F } _{ 1 }={ F } _{ 2 }={ F } _{ 3 }$

  2. ${ F }> _{ 1 }{ F } _{ 2 }{ >F } _{ 3 }$

  3. ${ F }> _{ 2 }{ F } _{ 3 }{ >F } _{ 1 }$

  4. ${ F } _{3}={ F } _{ 1 }{ >F } _{ 2 }$


Correct Option: D
Explanation:

Nuclear forces are the strong forces of attraction which hold together the nucleons (neutrons and protons) in the tiny nucleus of an atom, inspite of strong electrostatic forces of repulsion between protons. Nuclear forces act between a pair of neutrons, a pair of protons and also between a neutron, proton pair with the same strength. This shows that nuclear forces are independent of charge. 
The attractive nuclear force is the same for any pair of nucleons. Thus$ F _1 = F _3$ when there are no electrostatic forces, but $F _2 =$ attractive nuclear force - repulsive electrostatic force
Hence
$F _3 = F _1 > F _2$

The binding energy of $\alpha $-particle is ( if ${ m } _{ p }=1000785$ $u,{ m } _{ n }=1.00866$ u and ${ m } _{ \alpha  }=4.00274u$)

  1. $56.42 MeV$

  2. $2.821 MeV$

  3. $28.21 MeV$

  4. $32.4 MeV$


Correct Option: C

For a pair production, the minimum frequency of the gamma ray must be:

  1. 2.5 x 10$^{14}$ Hz

  2. 2.5 x 10$^{20}$ Hz

  3. 2.5 x 10$^{28}$ Hz

  4. 2.5 x 10$^{34}$ Hz


Correct Option: B
Explanation:

For production to occur, the energy must be greater than the rest mass of proton + electron.
$E > ( 2 \times 0.511\ MeV = 1.022\ MeV)$
$h \nu > 1.022 \times  10^{6} \times 1.6 \times 10^{-19} J$
$\nu > \dfrac{1.6352 \times  10^{-13}}{h} = 2.5 \times 10^{20} Hz$

The energy released when a positron is annihilated is

  1. $0.51 MeV$

  2. $0.58 MeV$

  3. $185 MeV$

  4. $200 MeV$


Correct Option: A
Explanation:

Positron, also called positive electron,  positively charged subatomic particle having the same mass and magnitude of charge as the electron.

mass of $e^+ = 9.11\times 10^{-31}Kg$

$E =mc^2=  9.11\times 10^{-31}\times (3\times 10^8)^2  J$

                 $= {0.51  \ MeV}$

The energy of annihilation of positron is ${0.51 \ MeV}$.

If the energy of an electron in Hydrogen atom is given by expression, $-1312 /{ n }^{ 2 }kJ{ mol }^{ -1 }$, then the energy required to excite the electron from ground state to second orbit is 

  1. $328 kJ{ mol }^{ -1 }$

  2. $656 kJ{ mol }^{ -1 }$

  3. $984 kJ{ mol }^{ -1 }$

  4. $1312 kJ{ mol }^{ -1 }$


Correct Option: C
Explanation:

$\begin{array}{l} E=-1312\left( { \frac { 1 }{ 4 } -1 } \right)  \ =\frac { { -1312\times 3 } }{ 4 }  \ =328\times 3 \ =984\, kJmo{ l^{ -1 } } \ Hence, \ option\, \, C\, \, is\, \, correct\, answer. \end{array}$