A) 240.8pm
B) 151.8pm
C) 75.5pm
D) 300.5pm
Correct Answer: B
Solution :
In case of body centred cubic (bcc) crystal, \[\frac{d[B{{r}_{2}}]}{dt}=-\frac{3}{5}\frac{d[B{{r}^{-}}]}{dt}\] Hence, atomic radius of lithium, \[\frac{d[B{{r}_{2}}]}{dt}=-\frac{5}{3}\frac{d[B{{r}^{-}}]}{dt}\] \[\frac{d[B{{r}_{2}}]}{dt}=\frac{5}{3}\frac{d[B{{r}^{-}}]}{dt}\] \[\frac{d[B{{r}_{2}}]}{dt}=\frac{3}{5}\frac{d[B{{r}^{-}}]}{dt}\]
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