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Chapter 7: Problem 53

An electron confined to a box of width 0.15 nm by infinite potential energy barriers emits a photon when it makes a transition from the first excited state to the ground state. Find the wavelength of the emitted photon.

Short Answer

Expert verified
The energy difference between the first excited state and the ground state is \(\Delta E = \frac{3h^2}{8mL^2}\), where \(h = 6.63 \times 10^{-34} Js\), \(m = 9.11 \times 10^{-31} kg\), and \(L = 0.15 \times 10^{-9} m\). After calculating \(\Delta E\), the wavelength of the emitted photon can be found using the energy-wavelength relation: \(\lambda = \frac{hc}{\Delta E}\), where \(c = 3 \times 10^8 m/s\).

Step by step solution

01

Find the energy difference between the first excited state and the ground state

To find the energy difference between the first excited state (n=2) and the ground state (n=1), we can subtract the energy of the ground state from the energy of the first excited state using the formula given: $$\Delta E = E_2 - E_1 = \dfrac{h^2(2^2)}{8mL^2} - \dfrac{h^2(1^2)}{8mL^2}$$ $$\Delta E = \dfrac{h^2(4-1)}{8mL^2} = \dfrac{3h^2}{8mL^2}$$ Now substitute the given values of constants and width of the box: - \(h = 6.63 \times 10^{-34} Js\) - \(m = 9.11 \times 10^{-31} kg\) - \(L = 0.15 \times 10^{-9} m\) $$\Delta E = \dfrac{3(6.63 \times 10^{-34})^2}{8(9.11 \times 10^{-31})(0.15 \times 10^{-9})^2}$$ Calculate the energy difference \(\Delta E\).
02

Calculate the wavelength of the emitted photon

Now that we have found the energy difference, we can use the energy-wavelength relation to find the wavelength of the emitted photon: $$\lambda = \dfrac{hc}{\Delta E}$$ Substitute the values of the constants and the value of \(\Delta E\) found in step 1: - \(h = 6.63 \times 10^{-34} Js\) - \(c = 3 \times 10^8 m/s\) $$\lambda = \dfrac{(6.63 \times 10^{-34})(3 \times 10^8)}{\Delta E}$$ Calculate the wavelength \(\lambda\). Now we have the wavelength of the emitted photon.

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Most popular questions from this chapter

Suppose an electron is confined to a region of length \(0.1 \mathrm{nm}\) (of the order of the size of a hydrogen atom) and its kinetic energy is equal to the ground state energy of the hydrogen atom in Bohr's model (13.6 eV). (a) What is the minimum uncertainty of its momentum? What fraction of its momentum is it? (b) What would the uncertainty in kinetic energy of this electron be if its momentum were equal to your answer in part (a)? What fraction of its kinetic energy is it?

A gas of helium atoms at \(273 \mathrm{K}\) is in a cubical container with \(25.0 \mathrm{cm}\) on a side. (a) What is the minimum uncertainty in momentum components of helium atoms? (b) What is the minimum uncertainty in velocity components? (c) Find the ratio of the uncertainties in (b) to the mean speed of an atom in each direction.

A 5.0-eV electron impacts on a barrier of with 0.60 nm. Find the probability of the electron to tunnel through the barrier if the barrier height is (a) \(7.0 \mathrm{eV} ;\) (b) \(9.0 \mathrm{eV} ;\) and (c) \(13.0 \mathrm{eV}\)

When an electron and a proton of the same kinetic energy encounter a potential barrier of the same height and width, which one of them will tunnel through the barrier more easily? Why?

A particle of mass \(m\) is confined to a box of width L. If the particle is in the first excited state, what are the probabilities of finding the particle in a region of width \(0.020 L\) around the given point \(x:\) (a) \(x=0.25 L ;\) (b) \(x=0.40 L ;(\mathrm{c}) x=0.75 L ;\) and (d) \(x=0.90 L\)
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