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Grade 11General Physics

The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV. (a) What is the kinetic energy of the electron in this state? (b) What is the potential energy of the electron in this state? (c) Which of the answers above would change if the choice of the zero of potential energy is changed?

Profile image of Jayant Kumar
12 Years agoGrade 11
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1 Answer

Profile image of Saurabh Koranglekar
6 Years ago

To understand the energy levels in a hydrogen atom, especially for the first excited state, we can break down the concepts of kinetic energy (KE) and potential energy (PE) associated with the electron. The total energy of the electron in the first excited state is known to be about –3.4 eV. Let’s analyze this step by step.

Determining Kinetic Energy

The total energy (E) of an electron in an atom is given by the sum of its kinetic energy and potential energy. For a hydrogen atom, the relationship can be described by the equation:

E = KE + PE

In the hydrogen atom, the potential energy is defined as:

PE = –(k * e²)/r

Where k is Coulomb’s constant, e is the elementary charge, and r is the radius of the electron's orbit. In this case, we can use the known relationships derived from Bohr's model of the hydrogen atom.

For the first excited state (n=2), the total energy is given by:

E = –13.6 eV/n²

Plugging in n=2:

E = –13.6 eV / 2² = –3.4 eV

In Bohr's model, the kinetic energy is related to the potential energy by the equation:

KE = -0.5 * PE

Thus, we can express the potential energy for this state:

PE = 2 * KE

Calculating the Kinetic Energy

Since we know that the total energy E = KE + PE, we can express KE in terms of the total energy:

KE = E - PE

Given that PE = 2 * KE, we can substitute to find:

  • KE = E + 2 * KE
  • Solving gives us KE = –(3.4 eV)/2 = –1.7 eV

Therefore, the kinetic energy of the electron in the first excited state is approximately **1.7 eV** (positive value, since energy is always expressed in magnitude).

Finding Potential Energy

Now, using the relationship mentioned earlier, we can find the potential energy:

  • Since PE = 2 * KE, we have:
  • PE = 2 * (–1.7 eV) = –3.4 eV

Thus, the potential energy of the electron in this state is approximately **–3.4 eV**.

Impact of Changing the Zero of Potential Energy

When discussing the zero of potential energy, it’s important to note that potential energy is relative. If we choose a different reference point for potential energy, both kinetic and potential energy values may change, but the total energy will remain the same.

For example, if we set the zero of potential energy at infinity (where the electron is free and has zero potential), the potential energy would change, leading to a new value for KE based on the new PE. However, the total energy, which is the sum of these two components, will still equal –3.4 eV, as it is a characteristic of the state the electron occupies.

In summary, in the first excited state of hydrogen:

  • Kinetic Energy (KE) is approximately 1.7 eV.
  • Potential Energy (PE) is approximately –3.4 eV.
Changing the zero of potential energy affects the individual values of KE and PE, but not the total energy of –3.4 eV.