If $(\frac{0.51 \times 10^{- 10}}{4})$ is the radius of smallest electron orbit in hydrogen like atom, then this atom is

hydrogen atom

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H e^{+}

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L i^{2 +}

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B e^{3 +}

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Solution

The correct option is D $B e^{3 +}$

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Similar questions

(\frac{0.51 \times 10^{- 10}}{4}) m

\frac{0.51 \times 10^{- 10}}{4}

4 R y d b e r g s

1 R y d b e r g = 2.2 \times 10^{- 18} J

Correct order of radius of the $1^{s t}$ orbit of $H$ , $H e^{+}, L i^{2 +}, B e^{3 +}$ is:

3

x

4

{L i}^{2 +}

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