The ground state energy of hydrogen atom is 13.6 eV. When its electron is in the first excited state, its excitation energy is:

10.2 eV

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

zero

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.4 eV

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6.8 eV

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 10.2 eV
The energy of an electron in Bohr’s orbit of hydrogen atom is given by the expression:

$E_{n} = - 13.6 \frac{Z^{2}}{n^{2}}$ ; [where n = energy state of electron]

For hydrogen Z = 1

$E_{1} = - 13.6 e V$ (given)

For the first excited state, n = 2,

Therefore, $E_{2} = \frac{- 13.6}{2^{2}} e V$

$E_{2} = - 3.4 e V$

Now, $Δ E = E_{2} - E_{1}$

$= - 3.4 - (- 13.6) = 10.2 e V$

Suggest Corrections

Similar questions

Q. In a hydrogen atom, if the energy of an electron in ground state is

- 13.6 eV

, then its energy in the 2nd excited state is

Q. Total energy of an electron in the ground state of hydrogen atom is

- 13.6 e V

. Its total energy, when hydrogen atom is in the first excited state, is:

Q. The ground state energy of hydrogen atom is

- 13.6

eV. The energy of second excited state

H e^{+} i o n

in eV is:

Excitation energy of a hydrogen like ion in its first excitation state is 40.8 eV. Energy needed to remove the electron from the ion in ground state is

Q. In a hydrogen atom, if the energy of an electron in ground state is

- 13.6 eV

, then its energy in the 2nd excited state is

Join BYJU'S Learning Program