Q69 / 90 · JEE Main 2019 (09 Apr Shift 2) · Physics

A $H e^{+}$ ion is in its first excited state. Its ionization energy is:

Correct answer
A
$13.60 e V$
B
$48.36 e V$
C
$54.40 e V$
D
$6.04 e V$

Solution

E = \frac{- (13.6) z^{2}}{n^{2}}

= \frac{- (13.6) {(2)}^{2}}{2^{2}} = - 13.60 e V

Total energy of

H e^{+}

ion is in its first excited state is

- 13.60 e V

. Hence, its ionization energy will be

13.60 e V

Hence, the correct answer is option: 'A'.

Relevant Questions

The ratio of momentum of the photons of the and line of Balmer series of Hydrogen atoms is . The possible values of and are:-06 Apr, 2026 (M)Bohr's Atomic Model
In the hydrogen atom, the electron makes a transition from the higher orbit ( ) to a lower orbit ( ). The ratio of the radius of the orbits 05 Apr, 2026 (M)Bohr's Atomic Model
Angular momentum of an electron in a hydrogen atom is , then the energy of the electron is _______ eV.02 Apr, 2026 (M)Bohr's Atomic Model
Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the and orbits of hydrogen at02 Apr, 2026 (E)Bohr's Atomic Model
Two electrons are moving in orbits of two hydrogen like atoms with speeds and respectively. If the radii of these orbits are nearly same the24 Jan, 2026 (M)Bohr's Atomic Model
The energy of an electron in an orbit of the Bohr's atom is where is the ground state energy. If is the angular momentum of the electron in 21 Jan, 2026 (E)Bohr's Atomic Model