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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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