In hydrogen atom, energy of first excited state is - 3.4ev, then the K.E. of same orbital of hydrogen atom is ____evCorrect answer is '3.4'. Can you explain this answer?

Question

Answer

Explanation:

The energy levels of an electron in a hydrogen atom are quantized, meaning they can only have certain specific values. These energy levels are given by the equation:

E = -13.6 eV / n^2

where E is the energy of the level and n is the principal quantum number. The energy of the first excited state is given as -3.4 eV, which corresponds to n = 2.

Calculating the Kinetic Energy:

The total energy of the electron in an orbital can be given as the sum of its kinetic energy (KE) and potential energy (PE). In the case of a hydrogen atom, the potential energy is negative and equal in magnitude to the total energy.

Total Energy (E) = KE + PE

Since the potential energy is equal to the total energy, we can rewrite the equation as:

E = KE + (-E)

Simplifying this equation, we get:

KE = -E - (-E)
= -E + E
= 0

Therefore, the kinetic energy of the electron in the first excited state of a hydrogen atom is 0 eV.

However, it's important to note that this result is specific to the first excited state. In general, the kinetic energy of an electron in a hydrogen atom can take on different values depending on the energy level it occupies.

Answer

Total energy = 1/2 potential energy
potential energy=2(-3.4)
kinetic energy=Total energy-potental energy=
-3.4-(-6.8)=3.4

Answer

+3.4eV

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