In the $n^{th}$ orbit, the energy of an electron is $E_{n} = \frac{13.6}{n^{2}} eV$ for a hydrogen atom. The energy required to take the electron from first orbit to second orbit will be,

10.2 eV

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12.1 eV

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13.6 eV

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3.4 eV

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Solution

The correct option is A $10.2 eV$
We know, $E_{n} = \frac{- 13.6}{n^{2}}$
where $n = 1, 2, 3, . . . . .$
For $n = 2$ , $E_{2} = - \frac{13.6}{2^{2}} = - 3.4 eV$
For $n = 1$ , $E_{1} = - \frac{13.6}{1^{2}} = - 13.6 eV$
So, the energy required to take electron from first to second orbit is,
$E_{1 \to 2} = E_{2} - E_{1} = - 3.4 - (- 13.6) = 10.2 eV$

Hence, $(A)$ is the correct answer.

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In the nth orbit, the energy of an electron is En=13.6n2 eV for a hydrogen atom. The energy required to take the electron from first orbit to second orbit will be,

The correct option is A 10.2 eVWe know, En=−13.6n2 where n=1,2,3,..... For n=2, E2=−13.622=−3.4 eV For n=1, E1=−13.612=−13.6 eV So, the energy required to take electron from first to second orbit is, E1→2=E2−E1=−3.4−(−13.6)=10.2 eV Hence, (A) is the correct answer.

In the $n^{th}$ orbit, the energy of an electron is $E_{n} = \frac{13.6}{n^{2}} eV$ for a hydrogen atom. The energy required to take the electron from first orbit to second orbit will be,