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Chapter 37: Problem 27

What is the ratio of energy difference between the ground state and the first excited state for an infinite square well of length \(L\) to that of length \(2 L\). That is, find \(\left(E_{2}-E_{1}\right)_{L} /\left(E_{2}-E_{1}\right)_{2 L}\).

Short Answer

Expert verified
Answer: The ratio of the energy difference between the ground state and the first excited state for an infinite square well of length L to that of length 2L is 1/4.

Step by step solution

01

Find Energy of the Ground State and First Excited State for length L

To find the energy of the ground state (\(E_1\)) for length \(L\), we use the formula above with \(n = 1\). Similarly, to find the energy of the first excited state (\(E_2\)), we use \(n = 2\): $$E_{1}=\dfrac{1^2\hbar^2 \pi^2}{2mL^2}$$ $$E_{2}=\dfrac{2^2\hbar^2 \pi^2}{2mL^2}$$
02

Calculate the Energy Difference for Length L

Now that we have the energies of both states, the energy difference is given by: $$\Delta E_{L}=E_{2}-E_{1}$$ $$\Delta E_{L}=\dfrac{4\hbar^2 \pi^2}{2mL^2}-\dfrac{1\hbar^2 \pi^2}{2mL^2}$$ $$\Delta E_{L}=\dfrac{3\hbar^2 \pi^2}{2mL^2}$$
03

Find Energy of the Ground State and First Excited State for length 2L

Similar to Step 1, we find the energies for length \(2L\) as: $$E_{1}=\dfrac{1^2\hbar^2 \pi^2}{2m(2L)^2}$$ $$E_{2}=\dfrac{2^2\hbar^2 \pi^2}{2m(2L)^2}$$
04

Calculate the Energy Difference for Length 2L

Using the energies from Step 3, we calculate the energy difference: $$\Delta E_{2L}=E_{2}-E_{1}$$ $$\Delta E_{2L}=\dfrac{4\hbar^2 \pi^2}{2m(2L)^2}-\dfrac{1\hbar^2 \pi^2}{2m(2L)^2}$$ $$\Delta E_{2L}=\dfrac{3\hbar^2 \pi^2}{8mL^2}$$
05

Calculate the Desired Ratio

Finally, we calculate the ratio of the energy differences: $$\dfrac{\Delta E_{L}}{\Delta E_{2L}}=\dfrac{\frac{3\hbar^2 \pi^2}{2mL^2}}{\frac{3\hbar^2 \pi^2}{8mL^2}}$$ Using the fact that $ \dfrac{a}{c} / \dfrac{b}{c}=\dfrac{a}{b}$, $$\dfrac{\Delta E_{L}}{\Delta E_{2L}}=\dfrac{2mL^2}{8mL^2}=\dfrac{1}{4}$$ Therefore, the ratio of the energy difference between the ground state and the first excited state for an infinite square well of length \(L\) to that of length \(2L\) is \(\dfrac{1}{4}\).

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Most popular questions from this chapter

Find the probability of finding an electron trapped in a one-dimensional infinite well of width \(2.00 \mathrm{nm}\) in the \(n=2\) state between 0.800 and \(0.900 \mathrm{nm}\) (assume that the left edge of the well is at \(x=0\) and the right edge is at \(x=2.00 \mathrm{nm}\) ).

Consider a water vapor molecule in a room \(4.00 \mathrm{~m} \times\) \(10.0 \mathrm{~m} \times 10.0 \mathrm{~m}\). a) What is the ground state energy of this molecule, treating it as a simple particle in a box? b) Compare this energy to the average thermal energy of such a molecule, taking the temperature to be \(300 . \mathrm{K}\). c) What can you conclude from the two numbers you just calculated?

A particle of energy \(E=5 \mathrm{eV}\) approaches an energy barrier of height \(U=8 \mathrm{eV}\). Quantum mechanically there is a finite probability that the particle tunnels through the barrier. If the barrier height is slowly decreased, the probability that the particle will reflect from the barrier will a) decrease. b) increase. c) not change.

An experimental measurement of the energy levels of a hydrogen molecule, \(\mathrm{H}_{2}\), shows that the energy levels are evenly spaced and separated by about \(9 \cdot 10^{-20} \mathrm{~J}\). A reasonable model of one of the hydrogen atoms would then seem to be that of a hydrogen atom in a simple harmonic oscillator potential. Assuming that the hydrogen atom is attached by a spring with a spring constant \(k\) to the molecule, what is the spring constant \(k\) ?

An electron in a harmonic potential well emits a photon with a wavelength of \(360 \mathrm{nm}\) as it undergoes a \(3 \rightarrow 1\) quantum jump. What wavelength photon is emitted in a \(3 \rightarrow 2\) quantum jump? (Hint: The energy of the photon is equal to the energy difference between the initial and the final state of the electron.)
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