Tag: nuclei

Questions Related to nuclei

In the nuclear reaction ; $ _{92}U^{238}\rightarrow _{z}Th^{A}+ _{2}He^{4}$ the values of A and Z are: 

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. A=230,Z=8

  2. A=234,Z=90

  3. A=228, Z=94

  4. A=232, Z=

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Correct Option: B

Out side a nucleus 

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. Neutron is stable

  2. Proton and neuron both are stable

  3. Neutrons is unstable

  4. Neither neutrons nor proton is stable

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Correct Option: C

Consider a hypothetical annihilation of a stationary electron with a stationary positron. What is the wavelength of resulting radiation?

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. $\dfrac{h}{m _{0}c}$

  2. $\dfrac{h}{2m _{0}c}$

  3. $\dfrac{2h}{m _{0}c}$

  4. $\dfrac{h}{4\pi m _{0}c}$

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Correct Option: A
Explanation:
According to De Broglie eave length
$\lambda =\dfrac { h }{ mc } \quad \quad \quad E=\dfrac { hc }{ \lambda  } $
                             $\lambda =\dfrac { hc }{ E } =hc$
When an electron and a positron collide then a hypothetical annihilation of a stationary electron with a stationary position.

The atomic mass of $7 ^ { N ^ { 15 } }$ is 15.000108 a.m.u. and that  is of $8 ^ { \bigcirc ^ { 16 } }$ 15.994915 a.m.u. If the mass of a proton is 1.007825 a.m.u. then the minimum energy provided to remove the least tightly bound proton is

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. 0.0130181 MeV

  2. 12.13 MeV

  3. 13.018 MeV

  4. 12.13 eV

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Correct Option: B
Explanation:

$\begin{array}{l} \left( { { M _{ n } }+{ M _{ H } }-{ M _{ 0 } } } \right) \times 431.5 \ =\left[ { 15.000108+1.007827-15.994915 } \right] \times 431 \ =12.13\, Mev \end{array}$

The energy of the reaction ${ Li }^{ 7 }+p\longrightarrow 2{ He }^{ 4 }$ is (the binding energy per nucleon in ${ Li }^{ 7 }$ and ${ He }^{ 4 }$ nuclei are 5.60 and 7.06 MeV respectively.)

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. 17.3 MeV`

  2. 1.73 MeV

  3. 1.46 MeV

  4. Depends on binding energy of proton

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Correct Option: D
Explanation:

The energy released will depend on the energy equivalent of proton, taking its mass to be $1amu$ or
 corresponding energy as $931.5Mev$
we get energy released as $Q=E _{reactant} -E _{product}=7\times 5.6 +931.5 -2\times4\times  7.06=914.22Mev$

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

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. 23.6 MeV

  2. 26.9 MeV

  3. 13.9 MeV

  4. 19.2 MeV

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Correct Option: A

Find the  binding energy of a H atom in the state n = 2

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. 2.1 eV

  2. 3.4 eV

  3. 4.2 eV

  4. 2.8 eV

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Correct Option: B

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

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. $23.6 MeV$

  2. $26.9 MeV$

  3. $13.9 MeV$

  4. $19.2 MeV$

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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

nuclear reactions nuclear structure nuclei atomic nuclei physics

  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.

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Correct Option: A

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

nuclear reactions nuclear structure nuclei atomic nuclei physics

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

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

  3. $1863\ J$

  4. $931.5 \Mev$

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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$