In the $n^{t h}$ orbit, the energy of an electron $E_{n} = - \frac{13.6}{n^{2}} eV$ for 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 $10.2 eV$
$E_{1 \to 2} = - 3.4 - (- 13.6) = + 10.2 eV$

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

The correct option is A 10.2 eVE1→2=−3.4−(−13.6)=+10.2 eV

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

The correct option is A $10.2 eV$
$E_{1 \to 2} = - 3.4 - (- 13.6) = + 10.2 eV$