Question

In the hydrogen atom energy of the first excited state is -3.4eV. Then find out the $K.E$ of the same orbit of a hydrogen atom.
(A) $+3.4eV$
(B) $+6.8eV$
(C) $-13.6eV$
(D) $+13.6eV$

Answer 1

Hint: To find the kinetic energy you should know that, Total energy= -Kinetic energy. Energy of first excited state of hydrogen atom means n=2

Complete step by step solution:
As you have read in your chemistry lesson about the hydrogen spectrum where you have studied about the different energy levels exhibited by hydrogen.
Like n= (1, 2, 3, 4, 5)
Where n=1 represent the ground state,
n=2 represent the first excited state and so on.
- To find the energy of hydrogen atom of any level we have the formula,

\[|{{E}_{n}}|=\,\dfrac{-13.6}{{{n}^{2}}}eV\]……………… (1)
So in the question it is given that the energy of a hydrogen atom in its first excited state is$-3.4eV$.
-You can also check this by the formula (1),
 Here n=2, first excited state ,after putting the value of n in equation (1) you will get the same value of energy as given in question,
\[|{{E}_{n}}|=\,\dfrac{-13.6}{{{2}^{2}}}=\,-3.4eV\]
- Now we have to find the kinetic energy of the same orbit of the hydrogen atom;
-To find the kinetic energy we have the formula as,
 $\therefore $ Total energy= Kinetic energy + Potential energy………….(2)
- Where kinetic energy of electron =$\dfrac{kz{{e}^{2}}}{2r}$
Potential energy of electron =$-\dfrac{kz{{e}^{2}}}{r}$
Thus, $P.E\,=\,-2(K.E)$
By putting these value in equation (2) you will get,
Total energy = $-K.E$……… (3)
-In the question total energy of first excited state is given as -3.4eV,
  Now put this value in equation three,
\[\therefore \,K.E\,=\,-(-3.4\,eV)\]
\[\]\[K.E\,=\,+3.4eV\]

Thus the correct option will be (A).

Note: Take care of the formula that you are using to solve the question, because we have two formulas of kinetic energy but here we have to use the formula required to calculate the K.E of hydrogen atom. Formula that we have use is,
Total Energy = P.E + K.E = -K.E