To determine the ratio of kinetic energy (KE) to potential energy (PE) of an electron in a Bohr orbit of a hydrogen-like atom, we can start by recalling some fundamental principles of the Bohr model. In this model, the electron moves in a circular orbit around the nucleus, and we can derive the relationship between its kinetic and potential energies.
Understanding Kinetic and Potential Energy in Bohr Orbits
In the context of a hydrogen-like atom, the kinetic energy of the electron can be expressed as:
- Kinetic Energy (KE): KE = (1/2)mv²
Where m is the mass of the electron and v is its velocity. According to the Bohr model, the centripetal force required to keep the electron in its circular orbit is provided by the electrostatic force of attraction between the positively charged nucleus and the negatively charged electron. This leads us to the expression for potential energy:
- Potential Energy (PE): PE = - (k * e²) / r
Here, k is Coulomb's constant, e is the charge of the electron, and r is the radius of the orbit. The negative sign indicates that the potential energy is lower when the electron is closer to the nucleus.
Deriving the Ratio
In a stable orbit, the centripetal force is equal to the electrostatic force:
From this equation, we can express the kinetic energy in terms of potential energy. Rearranging gives:
Now, substituting the expression for potential energy:
Now, we can find the ratio of kinetic energy to potential energy:
- KE/PE = [(k * e²) / (2r)] / [- (k * e²) / r]
When we simplify this expression, we find:
Final Thoughts
Thus, the ratio of kinetic energy to potential energy for an electron in a Bohr orbit of a hydrogen-like atom is -1/2. Therefore, the correct answer to your question is B) -1/2.
This relationship is significant because it highlights the balance of forces in atomic structures and helps us understand the stability of electron orbits in quantum mechanics.