/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 An electron collides with a fixe... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 66

An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its \(K E\). Consequently the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The minimum \(K E\) of colliding electron is (A) \(10.2 \mathrm{eV}\) (B) \(1.9 \mathrm{eV}\) (C) \(12.09 \mathrm{eV}\) (D) \(13.6 \mathrm{eV}\)

Short Answer

Expert verified
The minimum kinetic energy of the colliding electron is (A) \(10.2 \mathrm{eV}\).

Step by step solution

01

Calculate energy of hydrogen atom at different energy levels

First, let's find the energy of hydrogen atom at its ground state (n=1), at n=2 level, and at n=3 level using the formula mentioned above: Ground state (n=1): \(E_1 = -\frac{13.6 \, \mathrm{eV}}{1^2} = -13.6 \, \mathrm{eV}\) Second energy level (n=2): \(E_2 = -\frac{13.6 \, \mathrm{eV}}{2^2} = -3.4 \, \mathrm{eV}\) Third energy level (n=3): \(E_3 = -\frac{13.6 \, \mathrm{eV}}{3^2} = -\frac{13.6}{9} \, \mathrm{eV} = -1.51 \, \mathrm{eV}\)
02

Calculate energy difference between n=3 and n=2

Now we need to find the energy difference between the n=3 and n=2 levels: \(\Delta E_{32} = E_2 - E_3 = (-3.4 - (-1.51)) \, \mathrm{eV} = 3.4 + 1.51 \, \mathrm{eV} = 4.91 \, \mathrm{eV}\) The energy difference between the third and second energy levels is 4.91 eV.
03

Determine the minimum KE of the colliding electron

Since the hydrogen atom was initially in its ground state and goes to the n=3 level after getting excited, the colliding electron must have provided the necessary energy for this to happen. This energy corresponds to the difference between the n=3 and n=1 levels. So, the minimum KE of the colliding electron is: \(KE_{min} = E_3 - E_1 = (-1.51 - (-13.6)) \, \mathrm{eV} = 1.51 + 13.6 \, \mathrm{eV} = 15.11 \, \mathrm{eV}\) However, the hydrogen atom emits a photon of energy 4.91 eV (determined in Step 2) after becoming excited. In order for the colliding electron to lose all its KE, we must subtract this energy from the energy provided by the electron: \(KE_{min} = 15.11 \, \mathrm{eV} - 4.91 \, \mathrm{eV} = 10.2 \, \mathrm{eV}\) As per the given options, the minimum kinetic energy of the colliding electron is: (A) \(10.2 \mathrm{eV}\)

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Key Concepts

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

Energy Levels
In atomic physics, the concept of energy levels is fundamental to understanding how electrons are arranged in an atom and how they interact with energy. Every atom has defined energy states, often called energy levels or quantum levels, where its electrons reside. When an electron absorbs energy, it moves to a higher energy level, and when it releases energy, it falls back to a lower energy level.

For hydrogen, the simplest atom, these energy levels can be calculated using the formula:\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]where \( E_n \) is the energy of the nth level, and 13.6 eV is the ionization energy for hydrogen.

These energy states are crucial for understanding phenomena such as photon emission and absorption, as they determine the electronic transitions that an atom can undergo.
Kinetic Energy (KE)
Kinetic Energy (KE) is a measure of the energy an object possesses due to its motion. It is given by the formula:

\[ KE = \frac{1}{2}mv^2 \]where \( m \) is the mass of the object, and \( v \) is its velocity. In the context of atomic physics, especially when dealing with electrons, KE represents the energy transferred when one particle collides with another, causing it to move or become excited.

When a moving electron collides with a stationary atom, the electron's kinetic energy can be transferred to the atom, exciting it to a higher energy level.

For the hydrogen atom in this scenario, the electron's KE must be sufficient to move the atom from its ground state to a specified energy level, such as \( n=3 \), as determined from the difference in energy levels.
Photon Emission
Photon emission occurs when an excited electron in an atom drops from a higher energy level to a lower one. This process releases energy in the form of a photon, which is a quantum of light energy. The wavelength of this emitted photon is related to the energy difference between the two levels involved in the transition.

In the case of the Balmer series for hydrogen, photons are emitted as electrons transition from higher energy levels (like \( n=3 \) or above) back to \( n=2 \). The largest wavelength, or the least energetic photon, in this series results from the smallest energy jump, such as from \( n=3 \) to \( n=2 \).

Understanding photon emission is crucial when studying spectrum lines emitted by substances, as these lines are unique fingerprints of the atoms involved. This knowledge is essential in fields like spectroscopy, where light is analyzed to understand the composition and properties of matter.

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Most popular questions from this chapter

Assertion: Work function of aluminium is \(4.2 \mathrm{eV}\). If two photons of each of energy \(2.5 \mathrm{eV}\) strike on an electron of aluminium, the electron is not emitted. Reason: In photoelectric effect a single photon interacts with a single electron and electron is emitted only if energy of each of incident photon is greater than the work function. (A) \(\mathrm{A}\) (B) \(\mathrm{B}\) (C) \(\overline{\mathrm{C}}\) (D) D

In certain element the \(K\)-shell electron energy is \(-18.525 \mathrm{keV}\) and the \(L\)-shell electron energy is \(-3 \mathrm{keV}\). When an electron jumps from the \(L\)-shell to \(K\) shell, an \(x\)-ray photon is emitted. The wavelength of the emitted \(x\)-rays is (A) \(0.8 \AA\) (B) \(1 \AA\) (C) \(0.6 \AA\) (D) \(1.2 \AA\)

If \(M_{O}\) is the mass of an oxygen isotope \({ }_{8} \mathrm{O}^{17}, M_{P}\) and are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is (A) \(\left(M_{O}-17 M_{N}\right) c^{2}\) (B) \(\left(M_{O}-8 M_{P}\right) c^{2}\) (C) \(\left(M_{O}-8 M_{P}-9 M_{N}\right) c^{2}\) (D) \(M_{O} c^{2}\)

The energy, the magnitude of linear momentum and orbital radius of an electron in a hydrogen atom corresponding to the quantum number \(n\) are \(E, P\) and \(r\) respectively. Then according to Bohr's theory for hydrogen atom (A) \(E P r\) is proportional to \(\frac{1}{n}\). (B) \(\frac{P}{E}\) is proportional to \(n\). (C) \(E r\) is constant for all orbits. (D) \(P r\) is proportional to \(n\).

When a certain metallic surface is illuminated with monochromatic light of wavelength \(\lambda\), the stopping potential for photoelectric current is \(3 V_{0}\). When the same surface is illuminated with the light of wavelength \(2 \lambda\), the stopping potential is \(V_{0}\). The threshold wavelength for the surface for photoelectric effect is (A) \(\frac{4 \lambda}{3}\) (B) \(4 \lambda\) (C) \(6 \lambda\) (D) \(8 \lambda\)
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