Home
Class 11
CHEMISTRY
If radiation correcsponding to second li...

If radiation correcsponding to second line of “Balmer series” of `Li^(2+)` ion, knocked out electron from first excited state of H-atom, then kinetic energy of ejected electron would be:

A

2.55 eV

B

4.25 eV

C

11.25 eV

D

19.55 eV

Text Solution

Verified by Experts

The correct Answer is:
D

`E = 13.6 xx Z^2[1/(n_1^2)- 1/(n^2)] = 13.6xx3^2 [1/(2^2) - 1/(4^2)]`
Promotional Banner

Similar Questions

Explore conceptually related problems

In excited state a carbon atom gets its one of the 2s electrons to

The total energy of an electron in the first excited state of hydrogen atom is -3.4eV. What is the kinetic energy of the electron in this state ?

The energy of electron in the ground state of Hydrogen atom is -13. 6ev. The kinetic energy of the electron is the 4th orbit is

The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV. What is the kinetic energy of the electron in this state ?

The total energy of an electron in the first excited state of the hydrogen atom is -3.4eV. What is the potential energy of the electron in this state ?

The number of unpaired electrons present in the first excited state of chlorine atom is

The number of unpaired electrons present in the first excited state of chlorine atom is

The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV. What is the potential energy of the electron in this state ?

The ground state energy of hydrogen atom is - 13.6eV. The kinetic energy of the electron in this state is: