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Chapter 6: Problem 26

(a) A green laser pointer emits light with a wavelength of \(532 \mathrm{nm}\). What is the frequency of this light? (b) What is the energy of one of these photons? (c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to an upper excited state. When the electrons return to the ground state, they lose the excess energy in the form of \(532-\mathrm{nm}\) photons. What is the energy gap between the ground state and excited state in the laser material?

Short Answer

Expert verified
The frequency of the light emitted by the green laser pointer is \(f = \frac{3.00 \times 10^8\, m/s}{532 \times 10^{-9} m} \approx 5.63 \times 10^{14} Hz\). The energy of one photon is \(E = (6.63 \times 10^{-34}\, Js) * f \approx 3.73 \times 10^{-19} J\). The energy gap between the ground state and the excited state in the laser material is \(\Delta E = E \approx 3.73 \times 10^{-19} J\).

Step by step solution

01

Calculate the frequency of the light

To find the frequency (f) of the light, we can use the formula relating the speed of light (c), wavelength (λ), and frequency (f): \[c = λ * f\] Rearrange the formula to solve for frequency: \[f = \frac{c}{λ}\] The speed of light (c) is approximately \(3.00 \times 10^8 \, m/s\), and the wavelength (λ) of the light is \(532\, nm\), which can be converted to meters: \(\lambda = 532 * 10^{-9} m\). Now, substitute the values of c and λ into the frequency formula: \[f = \frac{3.00 \times 10^8\, m/s}{532 \times 10^{-9} m}\]
02

Calculate the energy of one photon

To find the energy (E) of one photon, we can use the formula relating Planck's constant (h) and frequency (f): \[E = h * f\] Planck's constant (h) is approximately \(6.63 \times 10^{-34}\, Js\). Use the frequency calculated in Step 1 to find the energy of one photon: \[E = (6.63 \times 10^{-34}\, Js) * f\]
03

Calculate the energy gap between ground state and excited state

The energy gap (ΔE) between the ground state and the excited state in the laser material is equal to the energy of one photon, as calculated in Step 2. So, the energy gap is: \[\Delta E = E\] Calculate the values for frequency (f), energy of one photon (E), and the energy gap between ground state and excited state (ΔE) using the formulas and constants provided.

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Most popular questions from this chapter

If a sample of calcium chloride is introduced into a nonluminous flame, the color of the flame turns to orange ("flame test"). The light is emitted because calcium atoms become excited; their return to the ground state results in light emission. (a) The wavelength of this emitted light is $422.7 \mathrm{nm} .\( Calculate its frequency. (b) What is the energy of \)1.00 \mathrm{~mol}$ of these photons (a mole of photons is called an Einstein)? (c) Calculate the energy gap between the excited and ground states for the calcium atom.

(a) Using Equation 6.5 , calculate the energy of an electron in the hydrogen atom when \(n=3\) and when \(n=6\). Calculate the wavelength of the radiation released when an electron moves from \(n=6\) to \(n=3 .(\mathbf{b})\) Is this line in the visible region of the electromagnetic spectrum?

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Indicate whether energy is emitted or absorbed when the following electronic transitions occur in hydrogen: (a) from \(n=2\) to \(n=3,(\mathbf{b})\) from an orbit of radius 0.529 to one of radius \(0.476 \mathrm{nm},(\mathbf{c})\) from the \(n=9\) to the \(n=6\) state.

The watt is the derived SI unit of power, the measure of energy per unit time: \(1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}\). A semiconductor laser in a DVD player has an output wavelength of \(650 \mathrm{nm}\) and a power level of $5.0 \mathrm{~mW}$. How many photons strike the DVD surface during the playing of a DVD 90 minutes in length?
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