The energy difference (in eV) between fourth and second orbits for $H$ atom is :

1.23

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10.20

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2.55

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5.10

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Solution

The correct option is C $2.55$
It is given that the quantisation rule followed in aliens' planet is as follows:
$m v r = \frac{n h}{4 π}$
The change in quantisation rule will not affect the energy difference.
The energy difference (in eV) is given by the expression $= E_{n} - E_{n}^{'} = - 13.6 \times [\frac{1}{n^{2}} - \frac{1}{n^{' 2}}]$
We need to find the energy difference between fourth and second orbits.
$E_{4} - E_{2} = - 13.6 [\frac{1}{4^{2}} - \frac{1}{2^{2}}]$
$E_{4} - E_{2} = - 13.6 \times [0.0625 - 0.25] = - 13.6 \times (- 0.1875) = 2.55$ eV
Hence, the energy difference between fourth and second orbits for $H$ atom is $2.55$ eV.

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The energy difference (in eV) between fourth and second orbits for H atom is :

The energy difference (in eV) between fourth and second orbits for $H$ atom is :