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Chapter 40: Q. 7 (page 1175)

Show that the penetration distance $η$ has units of $m$ .

Short Answer

Expert verified

The penetration distance $η$ has units of $m$ is given below in step.

Step by step solution

01

Given Information

We need to find the penetration distance $η$ has units of $m$ .

02

Explanation

The equation of penetration distance is given as:

$η = \frac{h}{\sqrt{2 m (U_{0} - E)}}$

The unit of role="math" localid="1650172801611" $h$ is $J . s$ . Its dimension can be $[h] = M \times L^{2} \times T^{- 1}$ .

The unit of $m$ is $k g$ , so dimension is $[m] = M$ .

The unit of $(U_{0} - E)$ is $J$ , then dimension is $(U_{0} - E) = M \times L^{2} \times T^{- 2}$

Therefore, the dimension of $η$ is:

$[h] = \frac{M \times L^{2} \times T^{- 1}}{\sqrt{M \times L^{2} \times T^{- 2}}} = \frac{M \times L^{2} \times T^{- 1}}{M \times L \times T^{- 1}} = L$

Then, $η$ has the dimension of length. So its unit is $m$ .

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Most popular questions from this chapter

A finite potential well has depth $U_{0} = 2.00 e V$ . What is the penetration distance for an electron with energy

(a) $0.50 e V$

(b) $1.00 e V$ and

(c) $1.50 e V$ ?

An electron in a rigid box absorbs light. The longest wavelength in the absorption spectrum is $600 n m$ . How long is the box?

A helium atom is in a finite potential well. The atom’s energy is $1.0 eV$ below $U_{0}$ . What is the atom’s penetration distance into the classically forbidden region?

The energy of an electron in a $2.00 eV$ deep potential well is $1.50 eV$ . At what distance into the classically forbidden region has the amplitude of the wave function decreased to $25 %$ of its value at the edge of the potential well?

The graph in FIGURE EX40.16 shows the potential-energy function U(x of a particle. Solution of the Schrödinger equation finds that the n=3 level has $E_{3} = 0.5 e V$ and that the n=6 level has $E_{6} = 2.0 e V$ .

a. Redraw this figure and add to it the energy lines for the n=3 and n=6 states.

b. Sketch the n=3 and n=6 wave functions. Show them as oscillating about the appropriate energy line.

See all solutions

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