If $r_{1}$ is the radius of the first orbit of hydrogen atom, then the radii of second, third and fourth orbits in term of $r_{1}$ are :

r_{1}^{2}, r_{1}^{3}, r_{1}^{4}

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4 r_{1}, 9 r_{1}, 16 r_{1}

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8 r_{1}, 27 r_{1}, 64 r_{1}

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2 r_{1}, 6 r_{1}, 8 r_{1}

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Solution

The correct option is B $4 r_{1}, 9 r_{1}, 16 r_{1}$
Radii of $n^{t h}$ orbit= $\frac{n^{2} r_{1}}{Z}$

$r_{1} =$ radius of first orbit of $H -$ atom

So, $r_{2} = \frac{2^{2} r_{1}}{1} = 4 r_{1}$

$r_{3} = \frac{3^{2} r_{1}}{1} = 9 r_{1}$

$r_{4} = \frac{4^{2} r_{1}}{1} = 16 r_{1}$

So, $4 r_{1}, 9 r_{1}, 16 r_{1}$ .

Hence, the correct option is B.

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(1 + r) (1 + r^{2}) (1 + r^{4}) (1 + r^{8}),

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(\frac{1}{r_{1}} + \frac{1}{r_{2}}) (\frac{1}{r_{2}} + \frac{1}{r_{3}}) (\frac{1}{r_{3}} + \frac{1}{r_{1}}) = \frac{64 R^{3}}{a^{2} b^{2} c^{2}}

1 + r + r^{2} + r^{3} + . . . . . . . + r^{n} = (1 + r) (1 + r^{2}) (1 + r^{4}) (1 + r^{8})

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