Problem 72
Question
In hydrogen atom, energy of first excited state is \(-3.4\) \(\mathrm{eV}\). The kinetic energy of the same orbit of hydrogen atom would be (a) \(+3.4 \mathrm{eV}\) (b) \(+6.8 \mathrm{eV}\) (c) \(-13.6 \mathrm{eV}\) (d) \(+13.6 \mathrm{eV}\)
Step-by-Step Solution
Verified Answer
The kinetic energy of the first excited state is \(+3.4 \text{ eV}\), option (a).
1Step 1: Identify the System
We are dealing with a hydrogen atom in the first excited state. The energy provided for this state is \(-3.4\text{ eV}\).
2Step 2: Understand the Relation Between Total Energy and Kinetic Energy
In Bohr's model of the hydrogen atom, the total energy of the electron in an orbit is the sum of its kinetic energy (KE) and potential energy (PE). This relation is given by:\[ E = KE + PE \]For a hydrogen atom, the kinetic energy is equal to the negative of the total energy:\[ KE = -E \]
3Step 3: Calculate the Kinetic Energy Based on Given Energy
Since the total energy of the first excited state is given as \(-3.4\text{ eV}\), substitute \(E\) in the equation for kinetic energy:\[ KE = -(-3.4) = +3.4 \text{ eV}\]
4Step 4: Choose the Correct Option
Based on the calculated kinetic energy, the correct option is (a) \(+3.4 \text{ eV}\).
Key Concepts
Hydrogen Atom Energy LevelsKinetic Energy CalculationsFirst Excited State Energy
Hydrogen Atom Energy Levels
In the study of hydrogen atoms, one of the fascinating aspects is understanding its energy levels. Each hydrogen atom has a set of energy levels where its electron can reside. These levels are quantized, meaning electrons can only exist in specific energy states. This concept was articulated by Niels Bohr in his model of the hydrogen atom. According to Bohr's model, when an electron moves from a lower energy level to a higher one, it absorbs photons, and when it falls back to a lower energy state, it emits photons.
In the ground state, or the lowest energy level, the electron is closest to the nucleus. This level is referred to as the n=1 level and has the energy of -13.6 ext{ eV}. The first excited state is the n=2 level where the electron possesses higher energy. For the hydrogen atom, this first excited state has an energy of -3.4 ext{ eV}. Thus, each move between energy levels involves discrete energy packets, called quanta.
In the ground state, or the lowest energy level, the electron is closest to the nucleus. This level is referred to as the n=1 level and has the energy of -13.6 ext{ eV}. The first excited state is the n=2 level where the electron possesses higher energy. For the hydrogen atom, this first excited state has an energy of -3.4 ext{ eV}. Thus, each move between energy levels involves discrete energy packets, called quanta.
Kinetic Energy Calculations
In Bohr's model, the total energy (E) of an electron in a hydrogen atom is composed of kinetic energy (KE) and potential energy (PE). This relationship is expressed as:
E = KE + PE
It is important to emphasize a particular feature of this model: the kinetic energy equals the negative of the total energy. This might seem counterintuitive, but it is rooted in the nature of attractive forces between an electron and a proton.
These calculations are crucial for predicting how much energy an electron has while orbiting the nucleus in a particular energy state. For the hydrogen atom, if the total energy in any state is given, like -13.6 ext{ eV} for the ground state, the kinetic energy can be determined using KE = -E, ensuring KE will be a positive number.
These calculations are crucial for predicting how much energy an electron has while orbiting the nucleus in a particular energy state. For the hydrogen atom, if the total energy in any state is given, like -13.6 ext{ eV} for the ground state, the kinetic energy can be determined using KE = -E, ensuring KE will be a positive number.
- In the ground state: KE = 13.6 ext{ eV}
- In the first excited state: given E = -3.4 ext{ eV}, KE = 3.4 ext{ eV}
First Excited State Energy
The first excited state of a hydrogen atom refers to the condition when the electron occupies the second lowest energy level. At this position, it has more energy compared to the ground state and is less tightly bound to the nucleus. The energy level of the first excited state is
-3.4 ext{ eV}, meaning less energy is necessary for the electron to remain in this state compared to the ground state.
Moving from the ground state to the first excited state requires absorbing energy, precisely 10.2 ext{ eV}. This difference reflects the jump in energy states within Bohr's model. Conversely, when an electron releases energy and drops to a lower energy state, it emits photons, contributing to the hydrogen emission spectra.
Understanding this excited state is fundamental in various applications:
Moving from the ground state to the first excited state requires absorbing energy, precisely 10.2 ext{ eV}. This difference reflects the jump in energy states within Bohr's model. Conversely, when an electron releases energy and drops to a lower energy state, it emits photons, contributing to the hydrogen emission spectra.
Understanding this excited state is fundamental in various applications:
- Explaining atomic spectra
- Predicting electron transitions
- Analyzing physical phenomena related to light absorption and emission
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