Question

the ratio of the radius of 2nd orbit of Li2+ to that of 3rd orbit of He+ is

Answer 1

Explanation:

this question is from bohr's atomic model check it out for more clear concept

Answer 2

The given question is we have to find the ratio of the radius of the 2nd orbit of Li2+ to that of the 3rd orbit of He+ is

The atomic number of lithium is 3 and helium is 2.

The radius of an orbit is directly proportional to the

$r \: = 0.53 \: x {10}^{ - 10} \times \frac{ {n}^{2} }{z}$

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

$sor \alpha \frac{ {n}^{2} }{z}$

$r \: {li}^{ + 2} \alpha (\frac{ {2}^{2} }{3} )$

The orbit of the lithium is 2nd orbit.

$r \: {he}^{ + } \alpha (\frac{ {3}^{2} }{2} )$

As we are asked to find the ratio of lithium and helium, we have to divide the radius of lithium by the helium.

$\frac{radius \: of \: {lithium}^{ + 2} }{radius \: of {helium}^{ + } } = \frac{ \frac{ ({2}^{2} }{3} )}{ (\frac{ {3}^{2} }{2}) }$

we have to simplify the above expression by the general method.

$= \frac{ \frac{4}{3} }{ \frac{9}{2} }$

The value will get multiplied.

$\frac{4}{3} \times \frac{2}{9}$

On multiplying we get

$\frac{8}{27}$

On taking the cube root we get,

$\frac{2}{3}$

Therefore, the ratio of the radius of the 2nd orbit of Li2+ to that of the 3rd orbit of He+ is 2:3

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