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What should be the ratio of energies of the electrons of the first orbits of ${Na}^{+ 10}$ & H ?

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11:1

1:121

121:1

1:11

answer is B.

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Detailed Solution

$H = - 13.6 ev$

$H like Species = \frac{13.6}{n^{2}} \times Z^{2}$

$For {Na}^{+ 10} r = \frac{- 13.6}{I^{2}} \times {(11)}^{2}$

$= - 1645.6$

${Na}^{+ 10} : H = - 1645.6 : - 13.6 = 121 : 1$

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