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The ratio of kinetic energy and potential energy of an electron in a Bohr orbit of a hydrogen like species is a. \(-1\) b. \(+1\) c. \(-1 / 2\) d. \(+1 / 2\)

Short Answer

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+1/2

Step by step solution

01

Understanding the Concept of Bohr's Model

In Bohr's model of the atom, the electron moves in a circular orbit around the nucleus. The centripetal force required for the circular motion is provided by the electrostatic force of attraction between the electron and nucleus.
02

Kinetic Energy Formula in Bohr’s Model

The kinetic energy (KE) of an electron in the orbit is given by the formula: \( KE = \frac{1}{2}m v^2 \). However, due to the balance of forces in the orbit, \( KE \) is also equal to \(-\frac{E}{2}\), where \( E \) is the total energy of the electron.
03

Potential Energy Formula in Bohr’s Model

The potential energy (PE) for the electron due to electrostatic attraction is given by: \( PE = -k \cdot \frac{e^2}{r} \), where \( k \) is Coulomb’s constant, \( e \) is the charge of the electron, and \( r \) is the radius of the orbit. Potential energy is negative as it indicates an attractive force.
04

Calculating the Ratio of Kinetic Energy to Potential Energy

From the relations \( KE = -\frac{E}{2} \) and \( PE = -E \), the ratio \( \frac{KE}{PE} \) can be calculated as follows: \[ \frac{KE}{PE} = \frac{-\frac{E}{2}}{-E} = \frac{1}{2} \]. Therefore, the ratio of kinetic energy to potential energy is \( +\frac{1}{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy in Bohr's Model
In Bohr's model of the atom, each electron revolves around the nucleus in a fixed circular orbit. This is similar to how planets orbit the sun. The kinetic energy (KE) of the electron is essential because it is responsible for keeping the electron in motion within its orbit. According to Bohr's theory, the kinetic energy is calculated using the formula:
  • \[ KE = \frac{1}{2}mv^2 \]
  • where \( m \) is the mass of the electron and \( v \) is its velocity.

The electrons experience two forces: the centripetal force that keeps them in orbit and the electrostatic force that attracts them to the nucleus. Due to the balance of these forces, the total energy \( E \) is shared such that the kinetic energy is half and oppositely directed to the total energy, i.e., \(-\frac{E}{2}\). This relationship makes it easier to calculate energies in Bohr's model and understand electron movement more deeply.
Potential Energy in Bohr's Model
The potential energy (PE) in Bohr's model is just as crucial as kinetic energy. It arises from the electrostatic forces between the negatively charged electron and the positively charged nucleus. The electrostatic force is an attractive force, meaning it acts to draw the electron closer to the nucleus. Therefore, the potential energy is negative, representing this attraction.
  • The potential energy is given by the formula: \( PE = -k \frac{e^2}{r} \)
  • In this formula, \( k \) is Coulomb’s constant, \( e \) is the charge of the electron, and \( r \) is the radius of the orbit.

The concept of negative potential energy can be tricky, but it's essentially saying the system has less energy when the electron is closer to the nucleus — a necessary condition for the stability of the atom.
Energy Ratio in Bohr's Orbit
In the context of Bohr's model, understanding the ratio of kinetic energy to potential energy enriches our comprehension of atomic stability. In any Bohr orbit for a hydrogen-like atom, the ratio of kinetic energy of the electron to its potential energy can actually explain why electrons maintain stable orbits rather than spiraling into the nucleus.
  • This ratio is calculated using the expressions for kinetic and potential energy:
  • \[ \frac{KE}{PE} = \frac{-\frac{E}{2}}{-E} = \frac{1}{2} \]

This positive ratio of \( \frac{1}{2} \) suggests that even though electrons are bound to the nucleus through attractive forces, their kinetic energy is crucial for countering this attraction sufficiently to maintain a stable orbit. This balance of energies highlights the elegance of Bohr's model in explaining atomic structure.

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Most popular questions from this chapter

In which of the following case would the probability of finding an electron residing in a dxy orbital be zero? a. \(\mathrm{xz}\) and \(\mathrm{yz}\)-planes b. \(\mathrm{xy}\) and \(\mathrm{yz}\)-planes c. Z-direction, \(\mathrm{yz}\) and \(\mathrm{xz}\)-planes d. \(\mathrm{xy}\) and \(\mathrm{xz}\)-planes

(A): Electronic configuration of \(\mathrm{K}\) (19) is \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2}\) \(2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}\) \((\mathbf{R}):\) Energy of \(4 \mathrm{~s}<3 \mathrm{~d}\) hence, \(4 \mathrm{~s}\) is filled before 3d as decided by Aufbau rule.

The number of nodal planes is a px orbital is a. 1 b. 2 c. 3 d. 0

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In the following questions, two statements (Assertion) \(\mathrm{A}\) and Reason (R) are given. Mark a. If \(\mathrm{A}\) and \(\mathrm{R}\) both are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\) b. If \(\mathrm{A}\) and \(\mathrm{R}\) both are correct but \(\mathrm{R}\) is not the correct explanation of A c. A is true but \(\mathrm{R}\) is false d. A is false but \(\mathrm{R}\) is true e. \(\mathrm{A}\) and \(\mathrm{R}\) both are false (A): The kinetic energy of the photo-electron ejected increases with increase in intensity of incident light. \((\mathbf{R}):\) Increase in intensity of incident light increases the rate of emission.

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