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

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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 eVThe energy in the first orbit is - 13.6eV n=2 --------------- E2=−13.6(2)2=−3.4 eV E1→2=−3.4−(13.6)=+10.2 eV

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