If radius of third shell of He+ion is x A then radius of second shell will be
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0
Explanation:
▪️Radius of the Bohr's orbit is given by,
▪️r=4π2mZe2h2n2
▪️Substituting values of h, m for helium ion and the electronic charge r may be calculated as
▪️r=Z0.529n2A0
▪️Where 0.529A0 is is the Bohr's radius a0=0.529
▪️Substituting Z=2, n=2
▪️r=2a0A0
Answered by
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radius is directly proportional to the square of number of shells
=> x = k.3²
=> k = x/3²
now shells = 2
so, radius = k.2²
= x.2²/3³
= 4x/9
=> x = k.3²
=> k = x/3²
now shells = 2
so, radius = k.2²
= x.2²/3³
= 4x/9
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