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Which of the following is true when a hydrogen atom is raised from the ground state to an excited state,
1. Both K.E. and P.E. increase
2. Both K.E. and P.E, decrease
3. The P.E. decreases and K.E. increases
4. The P.E. increases and K.E. decreases

1 Answer

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Correct Answer - Option 4 : The P.E. increases and K.E. decreases

Concept: 

When an electron revolves around the nucleus, it carries Kinetic energy as well as Potential energy.

While Revolving around the nucleus, its kinetic energy is given by

\(K.E. = \dfrac{Z^2me^4}{8n^2h^2\epsilon_0}\)

and Potential Energy is given by 

\(P.E. = -\dfrac{Z^2me^4}{4n^2h^2\epsilon_0}\)

Thus, P.E.= -2K.E.

Now, Total energy is the sum of total kinetic energy and total potential energy. 

i.e, TE = PE + KE 

 

Explanation: 

We know that, 

\(K.E. = \dfrac{Z^2me^4}{8n^2h^2\epsilon_0}\)

K.E. \(\propto \dfrac{1}{n^2}\) 

So, as the hydrogen atom is raised from the ground state to an excited state, n increases.

Hence, Kinetic energy decreases.

 

Also, Potential energy is given by

\(P.E. = -\dfrac{Z^2me^4}{4n^2h^2\epsilon_0}\) 

P.E.\(\propto -\dfrac{1}{n^2}\)

So, as the hydrogen atom is raised from the ground state to an excited state, n increases 

This decreases the negative value of Potential energy, i.e PE becomes less negative. 

Hence, Potential energy increases

 

Option (4) is the correct answer.

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