If 13.6 eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from its first excited state is

Question

If 13.6 eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from its first excited state is (A) 10.2 eV (B) 5.1 eV (C) 3.4 eV (D) 6.8 eV

Tardigrade Admin · Accepted Answer

The energy of an electron in hydrogen atom is E=-(13 . 6/n2)eV So, the energy required to remove an electron from n=2 state is |E|=(13 . 6/22)eV=(13 . 6/4)=3.4eV

Q. If $13.6 e V$ energy is required to ionize the hydrogen atom, then the energy required to remove an electron from its first excited state is

$10.2 e V$

$5.1 e V$

$3.4 e V$

$6.8 e V$

The energy of an electron in hydrogen atom is
$E = - \frac{13.6}{n ^{2}} e V$
So, the energy required to remove an electron from $n = 2$ state is
$∣ E ∣ = \frac{13.6}{2 ^{2}} e V = \frac{13.6}{4} = 3.4 e V$

Q. If 13.6eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from its first excited state is

10.2eV

5.1eV

3.4eV

6.8eV