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Group Displacement Law

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Question

According to the nuclear reaction $_{4} B e +_{2} H e^{4} \to_{6} C^{12} +_{0} n^{1}$ , mass number of (Be) atom is

A

4

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B

9

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C

7

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D

6

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Solution

The correct option is B 9
Equate atomic no. and mass no.

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Similar questions

Q. According to the nuclear reaction

_{4} B e +_{2} H e^{4} \to_{6} C^{12} +_{0} n^{1}

, mass number of (Be) atom is

_{4} B e^{9} +_{2} H e^{4} \to_{6} C^{12} +_{0} n^{1}

.
shows the bombardment of Berilium nucleus with a:

Q. In nuclear reaction

_{4} B e^{9} +_{2} H e^{4} \to_{6} c^{12}

+X, X will be :

Q. In the nuclear reaction

_{12} M g^{24} +_{2} H e^{4} =_{0} n^{1} + ?

The product nucleus is

Q. The energy released in the reaction

_{1} H^{2} +_{1} H^{3}

\to_{2} H e^{4} +_{0} n^{1}

- ------- -

_{0} n^{1} = 1.00867 a m u

_{1} H^{2} = 2.01410 a m u

_{1} H^{3} = 3.0160 a m u

_{2} H e^{4} = 4.0026 a m u

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