Question

The ratio of the kinetic energy to the total energy of an electron in Bohr orbit is

• A. 1 : -1
• B. - 1 : 1
• C. 1 : 2
• D. 2 : -1

Answer 1

The Correct Option is A

To determine the ratio of the kinetic energy to the total energy of an electron in a Bohr orbit, we need to review the fundamental concepts of atomic physics involving the Bohr model of the atom.

The Bohr model posits that the electron revolves around the nucleus in certain allowed orbits without radiating energy. The energies associated with these orbits can be calculated as follows:

1. Kinetic Energy (KE): In the Bohr model, the kinetic energy of the electron is given by the formula: KE = \frac{1}{2} m v^2, where m is the mass of the electron and v is the velocity of the electron.
2. Potential Energy (PE): The electrostatic potential energy of the electron with respect to the nucleus is given by: PE = -\frac{e^2}{4 \pi \varepsilon_0 r}, where e is the charge of the electron, \varepsilon_0 is the vacuum permittivity, and r is the radius of the Bohr orbit.

Here we note that the kinetic energy is positive and the potential energy is negative.

3. Total Energy (TE): The total energy of the electron in a Bohr orbit is the sum of its kinetic and potential energies:
   TE = KE + PE = \frac{1}{2} m v^2 - \frac{e^2}{4 \pi \varepsilon_0 r}

Now, from Bohr's model, the total energy of an electron in orbit is given by: TE = -\frac{1}{2} \frac{e^2}{4 \pi \varepsilon_0 r}

It can be shown that the kinetic energy is numerically equal to negative half of the potential energy: KE = -\frac{1}{2} PE.

Thus, the total energy (being the algebraic sum of KE and PE) results in: TE = KE + PE = KE - 2 \times KE = -KE.

Therefore, the ratio of kinetic energy (KE) to the total energy (TE) is \frac{KE}{TE} = \frac{KE}{-KE} = -1.

Based on this analysis, the ratio of the kinetic energy to the total energy is 1 : -1.

The correct answer is, therefore, 1 : -1.