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The atomic number of an element X is 92, if it is isobaric to \mathrm{Y_{94}^{238}} then, the number of neutrons present in X is 

Option: 1

146


Option: 2

144


Option: 3

148


Option: 4

152


Answers (1)

best_answer

\mathrm{Y_{94}^{238}} has mass number of 238 and is isobaric to X.

This means that the mass number of X is also 238.

Therefore, the number of neutrons present in X is (238-92) i.e. 146

Hence, the correct option is (1)

Posted by

Gautam harsolia

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