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Question

The element Lawrencium (256103Lr) is named after Ernest Lawrence, whose invention - the cyclotron - helped discover many radioactive nuclei. Consider now, the alpha decay of 256103Lr as:
256103LrabMd+α,
Where α represents the alpha particle given out. What are the values of a and b?


A

a=252; b=105

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B

a=252; b=101

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C

a=260; b=101

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D

a=260; b=105

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Solution

The correct option is B

a=252; b=101


We have to find a and b in the following equation:
256103LrabMd+α.
We know an α-particle is just a helium-4 nucleus, 42He. We can write:
256103LrabMd+42He. (1)
Since the number of protons on both sides of the equation should be the same, and the lower index indicates the atomic number (Z), from (1) we can say,
103 = b + 2
b=101.
If Z is the same, and the number of nucleons is same for the decaying nucleus and products, it means the mass number A should be the same too, no? We can write from(1),
256 = a + 4
a=252.


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