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Chapter 7: Problem 138

Consider the following approximate visible light spectrum: Barium emits light in the visible region of the spectrum. If each photon of light emitted from barium has an energy of \(3.59 \times\) \(10^{-19} \mathrm{~J}\), what color of visible light is emitted?

Short Answer

Expert verified
The emitted visible light from barium is green with a wavelength of 553 nm.

Step by step solution

01

Determine the energy of the photon

Given the information, each photon of light emitted from barium has an energy of \(3.59 \times 10^{-19}\) J.
02

Calculate the wavelength of the emitted light

To calculate the wavelength, use the Planck's formula, which relates the energy of a photon to its frequency: \[E = h \nu\] Where: - E = energy of the photon (\(3.59 \times 10^{-19}\) J) - \(\nu\) = frequency of the photon (in Hz) - h = Planck's constant (\(6.626 \times 10^{-34}\) Js) Since the speed of light (\(c\)) is given by the product of the wavelength (\(\lambda\)) and the frequency (\(\nu\)), the formula can be rewritten as: \[E = h \frac{c}{\lambda}\] Solve for the wavelength (\(\lambda\)): \[\lambda = \frac{hc}{E}\] Substitute the values of h, c, and E: \[\lambda = \frac{(6.626 \times 10^{-34}\ \text{Js})(3.00 \times 10^8\ \text{m/s})}{3.59 \times 10^{-19}\ \text{J}}\]
03

Solve for the wavelength

Perform the calculations to find the wavelength: \[\lambda = \frac{(6.626 \times 10^{-34}\ \text{Js})(3.00 \times 10^8\ \text{m/s})}{3.59 \times 10^{-19}\ \text{J}} = 5.53 \times 10^{-7}\ \text{m} = 553\ \text{nm}\]
04

Determine the color of the light

Compare the wavelength obtained in the previous step with the range of wavelengths associated with different colors of visible light. The approximate visible light spectrum (in nanometers) is: - Violet: 380 - 450 nm - Blue: 450 - 495 nm - Green: 495 - 570 nm - Yellow: 570 - 590 nm - Orange: 590 - 620 nm - Red: 620 - 750 nm Since the wavelength we calculated is 553 nm, it falls within the green color range.
05

Conclusion

Therefore, the emitted visible light from barium is green with a wavelength of 553 nm.

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Most popular questions from this chapter

Write equations corresponding to the following. a. the fourth ionization energy of Se b. the electron affinity of \(\mathrm{S}^{-}\) c. the electron affinity of \(\mathrm{Fe}^{3+}\) d. the ionization energy of \(\mathrm{Mg}\)

Give the name and formula of each of the binary compounds formed from the following elements. a. \(\mathrm{Li}\) and \(\mathrm{N}\) b. \(\mathrm{Na}\) and \(\mathrm{Br}\) c. \(\mathrm{K}\) and \(\mathrm{S}\)

Write the expected electron configurations for each of the following atoms: \(\mathrm{Sc}, \mathrm{Fe}, \mathrm{P}, \mathrm{Cs}\), Eu, \(\mathrm{Pt}, \mathrm{Xe}, \mathrm{Br}\).

An excited hydrogen atom with an electron in the \(n=5\) state emits light having a frequency of \(6.90 \times 10^{14} \mathrm{~s}^{-1}\). Determine the principal quantum level for the final state in this electronic transition.

One of the visible lines in the hydrogen emission spectrum corresponds to the \(n=6\) to \(n=2\) electronic transition. What color light is this transition? See Exercise 138 .
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