The potential energy of an electron in the first Bohr orbit in the $H e^{+}$ ion is:

- 13.6 eV

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-27.2 eV

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-54.4 eV

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-108.8 eV

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Solution

The correct option is C -54.4 eV
$E_{n} = - \frac{(218 \times 10^{- 18} J) Z^{2}}{n^{2}} a t o m^{- 1}$
For $H e^{+}, n = 1, Z = 2$
$E_{1} = \frac{- (2.18 \times 10^{- 18} J) 2^{2}}{12} = - 8.72 \times 10^{- 18} J$
$1.602 \times 10^{- 19} J o u l e = 1 e V$
$- 8.72 \times 10^{- 18} = \frac{1}{1.602 \times 10^{- 19} J} \times - 8.72 \times 10^{- 18} J e V$
$= - 5.44 \times 10^{- 18 + 19}$
$= - 54.4 e V$

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The energy of an electron in the first Bohr orbit of H atom is -13.6eV. The possible energy value(s) of the excited state(s) for electrons in Bohr orbits to hydrogen is(are)

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The potential energy of an electron in the first Bohr orbit in the He+ ion is:

The correct option is C -54.4 eVEn=−(218×10−18J)Z2n2atom−1For He+,n=1,Z=2E1=−(2.18×10−18J)2212=−8.72×10−18J1.602×10−19Joule=1eV−8.72×10−18=11.602×10−19J×−8.72×10−18JeV=−5.44×10−18+19=−54.4eV

The potential energy of an electron in the first Bohr orbit in the $H e^{+}$ ion is: