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Chapter 40: Q. 18 (page 1175)

Two adjacent energy levels of an electron in a harmonic potential well are known to be $2.0 e V$ and $2.8 e V$ . What is the spring constant of the potential well?

Short Answer

Expert verified

The spring constant for potential well= $1.4 N / m$

Step by step solution

01

Step 1. Given information

Energy level of quantum harmonic oscillator =

$E_{n} = (n - \frac{1}{2}) 鈩� 蝇$

Here,

$n =$ denotes the state,

$鈩�$ = reduced Planck's constant and

$蝇 =$ classical angular frequency.

Adjacent energy level=

$E_{n + 1} = ((n + 1) - \frac{1}{2}) 鈩� 蝇$

$= (n + \frac{1}{2}) 鈩� 蝇$

02

Step 2. For the angular frequency

$= 鈩� 蝇$

$蝇 = \frac{E_{n + 1} - E_{n}}{鈩�}$

$蝇 = \frac{2.8 eV - 2.0 eV}{1.05 脳 10^{- 34} J 路 s}$

$= \frac{1.28 脳 10^{- 19}}{1.05 脳 10^{- 34} J 路 s}$

width="126">=1.22脳1015s-1

03

Step 3. For spring constant

$蝇 = \sqrt{\frac{k}{m}}$

Here,

$k =$ spring constant of the potential well

$m =$ electron mass.

$\frac{k}{m} = 蝇^{2}$

$k = m 蝇^{2}$

$k = (9.1 脳 10^{- 31} kg) {(1.22 脳 10^{15} s^{- 1})}^{2}$

$= 1.354 N / m 鈮� 1.4 N / m$

Thus, the spring constant for potential well = $1.4 N / m$

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Most popular questions from this chapter

Four quantum particles, each with energy E, approach the potential-energy barriers seen in FIGURE Q40.8 from the left. Rank in order, from largest to smallest, the tunneling probabilities ${(P_{tunnel})}_{a} to {(P_{tunnel})}_{d}$

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) $1.50 e V$ ?

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

Sketch the $n = 8$ wave function for the potential energy shown in FIGURE EX40.14.

Use the data from Figure 40.24 to calculate the first three vibrational energy levels of a $C = O$ carbon-oxygen double bond.

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