An electron is revolving round the nucleus of `He^(+)` ion with speed `2.2 xx 10^(6) m//s`. The potential energy of the electron is (if atomic number of `He = 2, 1 eV = 16 xx 10^(-19) J`. Mass of electron `= 9 xx 10^(-31) kg` )

Options

A. `- 13.61 eV`
B. `- 6.8 eV`
C. `- 27.22 eV`
D. zero

Answer

C

Explanation

`KE = (1)/(2)m xx v^(2) = (1)/(2) xx 9 xx 10^(-31) xx (2.2 xx 10^(6))^(2)`
`KE = (1)/(2) xx (9 xx 10^(-31) xx (2.2 xx 10^(6))^(2))/(1.6 xx 10^(-19))`
`KE = 13.61 eV`
`PE = -2 xx KE = -2 xx 13.61 = -27.22 eV`