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When  _{3}Li^{7} nuclei are bombarded by protons, and the resultant nuclei are _{4}Be^{8} the emitted particles will be

  • Option 1)

    neutrons

  • Option 2)

    alpha particles

  • Option 3)

    Beta particles

  • Option 4)

    gamma photons

 

Answers (1)

best_answer

As we learnt in

Q value -

X+Y\rightarrow Z+Q

Q=(M_{x}+M_{y}-M_{z})C^{2}

     

- wherein

M_{x} and M_{y} are mass of reactant

M_{z}  is mass of product

 

 

 ^{7}_{3}Li+^{1}_{1}p\rightarrow^{8}_{4}Be+^{A}_{z}X

From conservation of atomic number z + 4 = 4   or    z = 0 

From conservation of mass number 7 + 1 = 8 + A     or    A = 0 

So the particle is a massless particle and hence it is gamma photons. 

Correct option is 4.


Option 1)

neutrons

This is an incorrect option.

Option 2)

alpha particles

This is an incorrect option.

Option 3)

Beta particles

This is an incorrect option.

Option 4)

gamma photons

This is the correct option.

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