First slide

Bohr Model of the Hydrogen atom.

Question

The total energy of an electron in an atom in an orbit is -3.4 eV. Its kinetic and potential energies are, respectively

Moderate

A

$3.4 eV, 3.4 eV$
B

$- 3.4 eV, - 3.4 eV$
C

$- 3.4 eV, - 6.8 eV$
D

$3.4 eV, - 6.8 eV$

Solution

$Totalenergyofelectronin n^{th} orbit, E_{n} = \frac{- 13.6 Z^{2}}{n^{2}} eV .$
$Kinetic energy of electron in n^{th} orbit, K . E . = \frac{13.6 Z^{2}}{n^{2}} eV Potential energy of electron in n^{th} orbit, P.E. = \frac{- 27.2 Z^{2}}{n^{2}} eV Thus, total energy of electron, E_{n} = - K . E . = \frac{P . E .}{2} ∴ K.E. = 3.4 eV [Given E_{n} = - 3.4 eV] P.E. = 2 \times (- 3.4) = - 6.8 eV .$

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