In a hydrogen atom, if the energy of electron in the ground state is - x eV, then that in the 2 text nd excited state of He +is

Question

In a hydrogen atom, if the energy of electron in the ground state is - x eV, then that in the 2 text nd excited state of He +is (A) - xeV (B) -(4/9) × eV (C) +2 x eV (D) -(9/4) × eV

Tardigrade Admin · Accepted Answer

E n =( E text ground state / n 2) × z 2 =-(x/(3)2) ×(2)2=-(4/9) × eV E text ground =x eV given n=3 because 2 text nd excited state z=2 because

Q. In a hydrogen atom, if the energy of electron in the ground state is $- x e V$ , then that in the $2^{nd}$ excited state of $H e^{+}$ is

$- x e V$

$- \frac{4}{9} \times e V$

$+ 2 x e V$

$- \frac{9}{4} \times e V$

$E_{n} = \frac{E _{ground state}}{n ^{2}} \times z^{2}$
$= - \frac{x}{( 3 ) ^{2}} \times (2)^{2} = - \frac{4}{9} \times e V$
$E_{ground} = x e V$ given
$n = 3$ because $2^{nd}$ excited state
$z = 2$ because

Q. In a hydrogen atom, if the energy of electron in the ground state is −xeV, then that in the 2nd excited state of He+is

−xeV

−94​×eV

+2xeV

−49​×eV