Correct Answer - Option 1 : 10.2eV

__CONCEPT____:__

- The energy of electrons in any orbit is given by:

\(E_n = -13.6 \frac{Z^2}{n^2} \,eV\)

Where n = principal quantum number and Z = atomic number

__CALCULATION__:

Given - n = 2 and For hydrogen (Z) = 1

- The energy of electrons in 2nd is given by:

\(⇒ E_2 = \frac{-13.6}{4} \,eV=-3.4 \,eV\)

- The energy of electrons in 1st is given by:

\(⇒ E_1 = \frac{-13.6}{1} \,eV=-13.6 \,eV\)

- The energy required for sending electron from 1st orbit to the 2nd orbit will be:

⇒ ΔE = E_{2} - E_{1}

⇒ ΔE = -3.4eV + 13.6eV = 10.2 eV