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Question
The radius of \[\ce{^27_13X}\] nucleus is R. The radius of \[\ce{^125_53Y}\] nucleus will be ______.
Options
`5/3R`
`(13/53)^{1"/"3}R`
`(5/3R)^{1"/"3}`
`(13/53R)^{1"/"3}`
Solution
The radius of \[\ce{^27_13X}\] nucleus is R. The radius of \[\ce{^125_53Y}\] nucleus will be `underlinebb(5/3R)`.
Explanation:
As given in the question,
\[\ce{^27_13X -> X}\] is aluminium as the atomic number of aluminium is 13 and the mass number is 27.
\[\ce{^125_53Y -> Y}\] is the tellurium as the atomic number of tellurium is 53 and the mass number is 125.
Radius of the nucleus, R = `R_0(A)^{1"/"3}`
So, `R ∝ A^{1"/"3}`
So, `R_{Al}/R_{Te} = sqrt(27/125)`
`R_{Te} = (125/27)^{1"/"3} xx R`
`R_{Te} = 5/3R`
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