Question

for which orbit in He+ion,the circumference is 26.5A°. a)2,b)3,c)4,d)16​

Answer 1

Answer:

option (b) 3 is the right ans ......

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Answer 2

(c) 4

Given,

that the circumference for $He^{+}$ ions is 26.5A°

The formula of circumference = $2\pi r$

Putting the formula,

$r= \frac{26.5* 7}{2*22}$

upon solving the above equation,

r= 4.21A°

As helium ions behave as a hydrogen atom,

so, the formula for the radius will be,

$r= 0.529*\frac{n^{2} }{2}$

where, z is the atomic number, and n is the orbit

so, for $He^{+}$ ions, the orbit with the above circumference is,

$4.21= 0.529* \frac{n^{2} }{2}$

$n^{2} = \frac{4.21*2}{0.529}$

$n^{2} = \frac{8.42}{0.529}$

$n^{2} = 15.9$

round off is 16

$n= 4$

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