The ground state energy of hydrogen atom is `-13.6 eV`. What is the potential energy of the electron in this state

Options

A. Zero
B. `2eV`
C. `1 eV`
D. `-27.2 eV`

Answer

D

Explanation

From the first postulate of Bohr's atom model.
`(mv^(2))/(r ) = (Kze^(2))/(r^(2))`
`rArr (1)/(2)mv^(2) = (1)/(2)(Kze^(2))/(r )`
i.e. `KE` of electron `= (1)/(2)mv^(2) = (KZe^(2))/(2r)…..(1)`
Potential due to the nucleus, in the orbit in which electron is revolving `= (KZe^(2))/(r )`
`:.` Potential energy of electron `=` potential `xx` change
`= (KZe)/(r )(-e) =- (KZe^(2))/(r ) ......(2)`
Total energy of electron in the orbit
`E = KE + PE`
`= (1)/(2)(KZe^(2))/(r ) - (KZe^(2))/(r )`
`=- (KZe^(2))/(2r) ....(3)`
This implies that
`PE = 2e`
Given, that in ground state of hydrogen, total enegry
`R =- 13.6 eV`
`:. PE =- 2 xx 13.6 =- 27.2 eV`