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Q.

If the ionization energy of He⁺ is 19.6 × 10^-18 J per atom then the energy of Be³⁺ ion in the second stationary state is:

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a

$- 4.9 \times 10^{- 18} J$

b

$- 11.025 \times 10^{- 18} J$

c

$- 44.1 \times 10^{- 18} J$

d

None of these

answer is D.

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Detailed Solution

E_H in first orbit $= \frac{- 19.6 \times 10^{- 18}}{4}$

$E_{B e^{3 +}}$ in second orbit

$= - (\frac{19.6 \times 10^{- 18}}{4}) \times \frac{16}{4}$

$= - 19.6 \times 10^{- 18} J$

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