Question

Potential energy is -27.2 eV in the second orbit of $H{e}^{+}$, then calculate double of total energy in the first excited state of hydrogen atom.

• A. - 13.6 eV
• B. - 54.4 eV
• C. - 6.8 eV
• D. - 27.2 eV

Answer 1

The correct option is C - 6.8 eV

Potential Energy is -27.2 eV in second orbit of $H{e}^{+}$ then calculate double of total energy in the first excited state of hydrogen atom - 6.8 eV.

Total energy is one half the potential energy.
Second orbit of $H{e}^{+}$ has a total energy of $-27.2eV.$

Hence, its total energy will be $\frac{-27.2}{2}=-13.6eV$

The energy is proportional to $\frac{Z^2}{n^2}$

$E_{2,H}=E_{2,He+}\times \frac{Z_H^2}{Z_{He+}^2}=-13.6\times \frac{1^2}{2^2}=-3.4eV$

The double of the total energy in the first excited state of hydrogen atom is $2\times (-3.4eV)=-6.8eV$