1.

The radius of the smallest electron orbit in the hydrogen like atom is (0.51 xx 10^(-10))/(4)m, then it is :

Answer»

Hydrogen atom
`He^(+)`
`Li^(++)`
`Be^(+++)`

Solution :For hydrogen like ion, the radius of the nth orbit is `r_(n)=(n^(2)/Z)r_0`
where `r_(0)=0.51 xx 10^(-10)m`
Here, `r_(n) =(0.51 xx 10^(-10))/(4)m`
In the GROUND state n=1
`(0.51 xx 10^(-10))/(4) =1^(2)/Z xx 0.51 xx 10^(-10)`
Z=4
Hence, the atom is triply ionised Beryllium.


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