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Bohr's Postulates

What would be...

Question

What would be the radius of second orbit of ${H e}^{+}$ ion?

1.058 ˚ A

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3.023 ˚ A

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2.068 ˚ A

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4.458 ˚ A

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Solution

The correct option is A $1.058 ˚ A$
Radius of the nth orbit, $r_{n} = 0.529 \frac{n^{2}}{Z}$ $A^{o}$
For $H e^{+}$ ion, $Z = 2$
Radius of 2nd orbit, $r_{2} = 0.529 \frac{2^{2}}{2}$
$⟹ r_{2} = 1.058$ $˚ A$

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