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How much energy does it take to ionize a hydrogen atom that is in its first excited state?

Short Answer

Expert verified
The energy to ionize a hydrogen atom that is in its first excited state is 3.4 eV.

Step by step solution

01

Calculate the initial energy

First, calculate the initial energy of the hydrogen atom in its first excited state. Using the formula \(E_n = -13.6/n^2 eV\) and replacing \(n\) with 2, you get \(E_2 = -13.6/2^2 = -3.4 eV\)
02

Calculate final energy state

The final energy state, which represents the ionized atom, is zero because once the electron is removed, there are no more energy levels for the electron to occupy.
03

Calculate ionization energy

Finally, subtract the initial energy from the final energy. Because the final energy is zero, calculation can be simplified to: Ionization energy = 0 - (- 3.4 eV) = 3.4 eV

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Key Concepts

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

Atomic Energy Levels
In hydrogen atoms, electrons exist in specific energy levels, often visualized as different orbits around the nucleus. These levels are determined by quantum mechanics and are denoted as "n," starting with the ground state at n=1. Each level corresponds to a certain amount of energy.
The energy levels become closer together as they move further from the nucleus. This happens because the energy difference between levels decreases with increasing n. The formula to calculate the energy of each level is given by:
  • \(E_n = -\frac{13.6}{n^2}\) eV
For example, when an electron is in the first excited state (n=2), it possesses a specific energy calculated using this formula. Recognizing these distinct energy levels helps us understand electron behavior and the energy changes involved when we transition between these states.
Excited State
An excited state in an atom occurs when an electron moves to a higher energy level than its normal ground state. This can happen through various processes, such as absorption of a photon or collision with another particle. When an electron is in an excited state, it is more energetic than when it is in the ground state.
For a hydrogen atom, if the electron is excited from the ground state to n=2, it will have an energy of -3.4 eV. This state is known as the first excited state. Electrons in excited states are not stable and will eventually return to a lower energy state.
The return to the ground state involves releasing energy, typically in the form of light or electromagnetic radiation. Understanding excited states is crucial for exploring phenomena like chemical reactions and the emission of light in various colors.
Electronic Transitions
Electronic transitions involve the movement of electrons between different energy levels within an atom. These transitions can absorb or release energy, depending on whether the electron is moving to a higher or lower energy level.
When moving to a more energetic (higher) level, electrons must absorb energy, often from photons. Conversely, when they drop to a lower level, they release energy.
The amount of energy involved in these transitions is quantized and corresponds to the energy difference between the two levels. For hydrogen, transitioning from n=2 (first excited state) to n=∞ (ionization) releases 3.4 eV of energy, as this is the amount needed to remove the electron completely.
  • Absorption transitions: electron moves up a level, energy absorbed
  • Emission transitions: electron moves down a level, energy released
Understanding these transitions is vital for grasping how atoms interact with light and other forms of energy.

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Most popular questions from this chapter

A beam of electrons is incident upon a gas of hydrogen atoms. a. What minimum speed must the electrons have to cause the emission of 656 nm light from the \(3 \rightarrow 2\) transition of hydrogen? b. Through what potential difference must the electrons be accelerated to have this speed?

A 100 W lightbulb emits about 5 W of visible light. (The other \(95 \mathrm{W}\) are emitted as infrared radiation or lost as heat to the surroundings.) The average wavelength of the visible light is about \(600 \mathrm{nm},\) so make the simplifying assumption that all the light has this wavelength. a. What is the frequency of the emitted light? b. How many visible-light photons does the bulb emit per second?

In the atom interferometer experiment of Figure \(39.14,\) lasercooling techniques were used to cool a dilute vapor of sodium atoms to a temperature of \(0.0010 \mathrm{K}=1.0 \mathrm{mK}\). The ultracold atoms passed through a series of collimating apertures to form the atomic beam you see entering the figure from the left. The standing light waves were created from a laser beam with a wavelength of \(590 \mathrm{nm}\) a. What is the rins speed \(v_{\text {rass }}\) of a sodium atom \((A=23)\) in a gas at temperature \(1.0 \mathrm{mK} ?\) b. By treating the laser beam as if it were a diffraction grating, calculate the first-order diffraction angle of a sodium atom traveling with the rms speed of part a. c. How far apart are points \(B\) and \(C\) if the second standing wave is \(10 \mathrm{cm}\) from the first? d. Because interference is observed between the two paths, each individual atom is apparently present at both point B and point \(C\). Describe, in your own words, what this experiment tells you about the nature of matter.

An FM radio station broadcasts with a power of \(10 \mathrm{kW}\) at a frequency of \(101 \mathrm{MHz}\) a. How many photons does the antenna emit each second? b. Should the broadcast be treated as an electromagnetic wave or discrete photons? Explain.

The allowed energies of a simple atom are \(0.00 \mathrm{eV}\) \(4.00 \mathrm{eV},\) and 6.00 eV. An electron traveling with a speed of \(1.30 \times 10^{6} \mathrm{m} / \mathrm{s}\) collides with the atom. Can the electron excite the atom to the \(n=2\) stationary state? The \(n=3\) stationary state? Explain.

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