Question

If an electron in a hydrogen atom has moved from n=1 to n=10 orbit, the potential energy of the system has:
(A) increased
(B) decreased
(C) remain unchanged
(D) become zero

Answer 1

Hint: For hydrogen, the energy of the Bohr orbit (\mathrm{U})=-13.6 \mathrm{x}\left(\mathrm{z}^{2} / \mathrm{n}^{2}\right) where n is the orbit and Z for hydrogen is 1. Whenever there is a change in orbit for an electron, there is a change in the energy of the system. Let us now discuss the change.

Complete step by step solution:
-In the Bohr model of the hydrogen atom we take the potential energy of the electron to be zero at infinity, so the potential energy becomes negative as the electron approaches the hydrogen atom.
-When the electron is present in any other stationary state(orbit) n, it is attracted to the nucleus and its energy is lowered.
-The energy of the electron in orbit n is thus larger in absolute value than zero, but is denoted by negative sign, since it is being lowered.
-The negative sign for the energy thus represents the relative stability of the state with reference to zero energy state n as infinity.
\begin{array}{l}
\text { -For hydrogen, } \mathrm{U}=-13.6 / \mathrm{n}^{2} \\
\text { Thus, for } \mathrm{n}=1, \mathrm{U}_{1}=-13.6 \mathrm{eV} \text { . } \\
\text { For } \mathrm{n}=10, \mathrm{U}_{2}=-0.136 \mathrm{eV} \text { . }
\end{array}
Thus after comparison ${U_2}$ is less negative, hence it increases.

Clearly, the correct answer is (A).

Note: Thus, it should be noted that when the orbit increases, the potential energy increases. Thus, if it is close to the nucleus it gets attracted, thus the energy is lowered.