Consider the following nuclear reactions-I- 147N-42He-178O-XII- 94Be-42H-126He-Y- Then-

The correct option is C X is a proton and Y is a neutronIn the first reaction, #160; no of protons on LHS is 7 +2 =9 ; the no protons on RHS is 8 M ...

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Standard X

Physics

Properties of Alpha, Beta and Gamma Radiations

Consider the ...

Question

Consider the following nuclear reactions.
I. $_{7}^{14} N +_{2}^{4} H e \to_{8}^{17} O + X$
II. $_{4}^{9} B e +_{2}^{4} H \to_{6}^{12} H e + Y$ . Then.

X and Y are both protons

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X and Y are both neutrons

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X is a proton and Y is a neutron

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X is a neutron and Y is a proton

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Solution

The correct option is C X is a proton and Y is a neutron
In the first reaction, no of protons on LHS is $7 + 2 = 9$ ; the no protons on RHS is $8$
Mass number on LHS is $14 + 4 = 18$
Mass number on RHS is $17$
Hence, the no of protons X will have is $9 - 8 = 1$ and the no of neutrons X
has is $18 - 17 = 1$
In the 2nd reaction, no of protons Y has is $4 + 2 - 6 = 0$ and no of neutron Y has is $9 + 4 - 12 = 1$
Hence, both X and Y have one neutron.
In case of X energy can be released equivalent of one neutron mass.
Hence, X is a proton and Y is a neutron.

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Consider the following nuclear reactions.I. 147N+42He→178O+XII. 94Be+42H→126He+Y. Then.

Consider the following nuclear reactions.
I. $_{7}^{14} N +_{2}^{4} H e \to_{8}^{17} O + X$
II. $_{4}^{9} B e +_{2}^{4} H \to_{6}^{12} H e + Y$ . Then.