Question

In hydrogen atom, energy of first excited state is -3.4 eV. Then, KE of same orbit of hydrogen atom :-

• A. +3.4 eV
• B. +6.8 eV
• C. -13.6 eV
• D. +13.6 eV

Hint:

Total energy \((E_n)\)= \(KE + PE\) 

Answer 1

The Correct Option is A

To determine the kinetic energy (KE) of the electron in the first excited state of a hydrogen atom, we must understand the relationship between potential energy (PE), kinetic energy, and the total energy in an atom.

In an atom, the total energy E of the electron is related to its potential energy PE and kinetic energy KE by the equation:

E = KE + PE

The potential energy for an electron in a hydrogen atom can be related to the total energy as:

PE = 2 \times E

However, since the potential energy is conventionally considered negative for bound states, we correct this to:

PE = -2 \times KE

Given that the total energy E of the first excited state of hydrogen is -3.4 \, \text{eV}, we use the equation:

E = KE + PE
E = KE - 2 \times KE
-3.4 \, \text{eV} = KE - 2 \times KE

Solving this gives:

-3.4 \, \text{eV} = -KE

Thus:

KE = 3.4 \, \text{eV}

Therefore, the kinetic energy of the electron in the first excited state is +3.4 \, \text{eV}.

The correct answer is: +3.4 eV