For hydrogen atom energy of an electron in first excited sta

Question

For hydrogen atom, energy of an electron in first excited state is -3.4 eV, K . E. of the same electron of hydrogen atom is x eV. Value of x is _________ x 10^-1 eV. (Nearest integer)

DivineJEE · Accepted Answer

To determine the kinetic energy (K.E.) of an electron in the first excited state of a hydrogen atom, we need to understand the relationship between the total energy, potential energy, and kinetic energy in an atom.

In a hydrogen atom, the total energy (E) of an electron in the nth state is given by:

E_n = - \frac{13.6}{n^2} \mathrm{~eV}

For the first excited state, $n = 2$ . So, plugging in the value:

E_2 = - \frac{13.6}{2^2} = - \frac{13.6}{4} = - 3.4 \mathrm{~eV}

This value represents the total energy (E) of the electron in the first excited state. According to the virial theorem for an electron in a Coulomb potential (as in a hydrogen atom), the kinetic energy (K.E.) is equal to the negative of the total energy:

\mathrm{K.E.} = - E

Substituting the total energy we calculated:

\mathrm{K.E.} = - (-3.4) = 3.4 \mathrm{~eV}

Now, we need to find the value of

x

in the form of

x \times 10^{-1} \mathrm{~eV}

.

3.4 \mathrm{~eV} = 34 \times 10^{-1} \mathrm{~eV}

Therefore, the value of

x

is 34.

Explanation

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