    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. 256103Lr→abMd+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 D a=252 ; b =101We have to find a and b in the following equation: 256103Lr→abMd+α We know an α-particle is just a helium-4 nucleus, 42He. We can write: 256103Lr→abMd+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 should be same too. We can write from (1), 256 = a + 4 ⇒a=252  Suggest Corrections  0      Similar questions  Related Videos     