The allowed energies of a simple atom are 0.00 eV, 4.00 eV, and 6.00 eV. An electron traveling with a speed of 1.30×106 m/s collides with the atom. Can the electron excite the atom to the n = 2 stationary state? The n = 3 stationary state? Explain.

Step 1: Calculate the kinetic energy of the electron before the collision using the formula for kinetic energy: K=12mv2, where m is the mass of the electron (9.11×10⁻³¹ kg) and v is its speed (1.30×10⁶ m/s). Step 2: Convert the calculated kinetic energy from joules to electron volts (eV) using the conversion factor: 1 eV = 1.602×10⁻¹⁹ J. Step 3: Determine the energy difference required to excite the atom to the n=2 stationary state. The energy difference is given by ΔE=E_{2}-E_{1}, where E_{2} is 4.00 eV and E_{1} is 0.00 eV. Step 4: Similarly, calculate the energy difference required to excite the atom to the n=3 stationary state. The energy difference is given by ΔE=E_{3}-E_{1}, where E_{3} is 6.00 eV and E_{1} is 0.00 eV. Step 5: Compare the kinetic energy of the electron (in eV) with the energy differences calculated in steps 3 and 4. If the kinetic energy is greater than or equal to the energy difference for a given state, the electron can excite the atom to that state. Otherwise, it cannot.

Ch 38: Quantization

Knight Calc - Physics for Scientists and Engineers 5th Edition

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Knight Calc 5th Edition

Ch 38: Quantization

Problem 26

Chapter 38, Problem 26

The allowed energies of a simple atom are 0.00 eV, 4.00 eV, and 6.00 eV. An electron traveling with a speed of 1.30×10⁶ m/s collides with the atom. Can the electron excite the atom to the n = 2 stationary state? The n = 3 stationary state? Explain.

Verified step by step guidance

Step 1: Calculate the kinetic energy of the electron before the collision using the formula for kinetic energy:

K = \frac{1}{2} m v^{2}

, where

m

is the mass of the electron (9.11×10⁻³¹ kg) and

v

is its speed (1.30×10⁶ m/s).

Step 2: Convert the calculated kinetic energy from joules to electron volts (eV) using the conversion factor: 1 eV = 1.602×10⁻¹⁹ J.

Step 3: Determine the energy difference required to excite the atom to the n=2 stationary state. The energy difference is given by

ΔE = E

₂-E₁, where

E

₂ is 4.00 eV and

E

₁ is 0.00 eV.

Step 4: Similarly, calculate the energy difference required to excite the atom to the n=3 stationary state. The energy difference is given by

ΔE = E

₃-E₁, where

E

₃ is 6.00 eV and

E

₁ is 0.00 eV.

Step 5: Compare the kinetic energy of the electron (in eV) with the energy differences calculated in steps 3 and 4. If the kinetic energy is greater than or equal to the energy difference for a given state, the electron can excite the atom to that state. Otherwise, it cannot.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Energy Levels in Atoms

In quantum mechanics, atoms have discrete energy levels, which are the allowed energies that electrons can occupy. For the given atom, the energy levels are 0.00 eV (ground state), 4.00 eV (n=2), and 6.00 eV (n=3). An electron can only excite an atom to a higher energy state if it has enough energy to match the difference between these levels.

Kinetic Energy of the Electron

The kinetic energy (KE) of an electron can be calculated using the formula KE = 0.5 * m * v^2, where m is the mass of the electron and v is its velocity. For the electron traveling at 1.30×10^6 m/s, its kinetic energy must be compared to the energy differences between the atom's energy levels to determine if it can excite the atom.

Excitation and Energy Conservation

Excitation occurs when an electron transfers energy to an atom, allowing it to move from a lower to a higher energy state. According to the conservation of energy, the energy of the incoming electron must equal the energy difference between the initial and final states of the atom. If the electron's kinetic energy is greater than or equal to the energy required for excitation, the transition can occur.

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