Chapter 40: Q. 5 (page 1175)

FIGURE EX $40.5$ is the probability density for an electron in a rigid box. What is the electron’s energy, in $e V$ ?

Short Answer

Expert verified

The electron’s energy, in $e V$ is $16.9 e V$ .

Step by step solution

Given Information

We need to find the electron’s energy, in $e V$ .

Simplify

The given graph of the probability density has three antinodes, which means that the wave function also has three antinodes. And we know that the graph of $ψ_{n} (x)$ has $(n)$ antinodes .). So, the wave function shown in the graph corresponds to the $n = 3$ state. But the energy of an electron in a rigid box is given by:

$E_{n} = n^{2} \frac{h^{2}}{8 m L^{2}}$

Notice that $L$ is given to be $(0.45 n m)$ from the graph, which is the maximum value of $x$

substitute the numerical values for the quantities in the equation above to find $E_{3}$

$E_{3} = \frac{9 \times {(6.626 \times 10^{- 34} J \cdot s)}^{2}}{8 \times (9.11 \times 10^{- 31} kg) \times {(0.45 \times 10^{- 9} m)}^{2}} = 2.7 \times 10^{- 18} J$

or in electronvolt

$\begin{matrix} E_{3} = 2.7 \times 10^{- 18} J \times \frac{1 eV}{1.6 \times 10^{- 19} J} \\ E_{3} = 16.9 eV \end{matrix}$

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FIGURE EX $40.5$ is the probability density for an electron in a rigid box. What is the electron’s energy, in $e V$ ?

Short Answer

Step by step solution

Given Information

Simplify

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FIGURE EX 40.5is the probability density for an electron in a rigid box. What is the electron’s energy, in eV?

Short Answer

Step by step solution

Given Information

Simplify

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Linear Momentum

Scientific Method Physics

Quantum Physics

Physics of Motion

Geometrical and Physical Optics

Electric Charge Field and Potential

Study anywhere. Anytime. Across all devices.

FIGURE EX $40.5$ is the probability density for an electron in a rigid box. What is the electron’s energy, in $e V$ ?