/*! 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 34 An electron collides with a fixe... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 34

An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its kinetic energy. Consequently the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The minimum kinetic energy 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 (B) \(1.9 \mathrm{eV}\).

Step by step solution

01

Identify the energy levels involved in the transition

In the Balmer series, the initial energy level (\(n_i\)) is 2. The excited state with the largest wavelength corresponds to the lowest energy level above \(n_i\), which is \(n_f = 3\).
02

Find the energy difference between the two levels

Now, we will find the energy difference between the two levels using their energy values (\(E_n = -13.6 \mathrm{eV} /n^2\)): \(E_{2} = -13.6 \mathrm{eV} / 2^2 = -3.4 \mathrm{eV}\) \(E_{3} = -13.6 \mathrm{eV} / 3^2 = -1.51 \mathrm{eV}\) \(\Delta E = E_3 - E_2 = -1.51 \mathrm{eV} - (-3.4 \mathrm{eV}) = 1.89 \mathrm{eV}\)
03

Convert energy difference to the minimum kinetic energy

The energy difference is also the energy of the photon corresponding to the largest wavelength in the Balmer series. Since the colliding electron loses all its kinetic energy, its minimum kinetic energy must be equal to the energy difference we found: Minimum kinetic energy = \(\Delta E = 1.89 \mathrm{eV}\) The correct answer is (B) \(1.9 \mathrm{eV}\).

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