The ionization energy of the hydrogen atom is 13.6 eV. The potential energy of the electron in n = 2 state of hydrogen atom is

Question

The ionization energy of the hydrogen atom is 13.6 eV. The potential energy of the electron in n = 2 state of hydrogen atom is (A) + 3.4 eV (B) - 3.4 eV (C) + 6.8 eV (D) - 6.8 eV

Tardigrade Admin · Accepted Answer

Given, En=13.6 e V Energy of an electron in n text th state En=(-13.6 z2 e V/n2) ∴ Energy of an electron in n=2 state So, E2= (18-6 z2/(2)2)=-3.4 e V P E =2 En=2 =2 ×(-3.4) ≈-6.8 eV

Q. The ionization energy of the hydrogen atom is $13.6 e V$ . The potential energy of the electron in $n = 2$ state of hydrogen atom is

$+ 3.4 e V$

$- 3.4 e V$

$+ 6.8 e V$

$- 6.8 e V$

Given, $E_{n} = 13.6 e V$
Energy of an electron in $n^{th}$ state
$E_{n} = \frac{- 13.6 z ^{2} e V}{n ^{2}}$
$∴$ Energy of an electron in $n = 2$ state So,
$E_{2} = \frac{18 - 6 z ^{2}}{( 2 ) ^{2}} = - 3.4 e V$
$PE = 2 E_{n = 2}$
$= 2 \times (- 3.4)$
$\approx - 6.8 e V$

Q. The ionization energy of the hydrogen atom is 13.6eV. The potential energy of the electron in n=2 state of hydrogen atom is

+3.4eV

−3.4eV

+6.8eV

−6.8eV