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Chapter 4: Problem 132

Electromagnetic radiation emitted by magnesium has a wavelength of \(285.2 \mathrm{~nm}\). (a) Is this radiation visible to the eye? (b) What is the energy of this radiation?

Short Answer

Expert verified
The radiation with a wavelength of 285.2 nm emitted by magnesium is not visible to the eye, since the visible light range is approximately from 380 nm to 750 nm. The energy of this radiation is approximately \(6.97 × 10^{-19} J\).

Step by step solution

01

Calculate the frequency of the radiation

We are given the wavelength, λ, of the electromagnetic radiation as 285.2 nm. We will need to find the frequency, ν, of the radiation. We can do this using the speed of light, c, and the equation: \(c = λν\) The speed of light is approximately \(3.00 × 10^8 m/s\). First, convert the wavelength to meters: \(285.2\,\mathrm{nm} × \frac{1\,\mathrm{m}}{10^9 \,\mathrm{nm}} = 2.852 × 10^{-7}\, \mathrm{m}\) Now, we can rearrange the equation to solve for the frequency: \(ν = \frac{c}{λ}\)
02

Find the frequency of the radiation

Plug in the values of the speed of light (c) and the wavelength (λ) into the equation: \(ν = \frac{3.00 × 10^8 m/s}{2.852 × 10^{-7} m} \approx 1.052 × 10^{15} Hz\) So, the frequency of the electromagnetic radiation is approximately \(1.052 × 10^{15} Hz\).
03

Determine if the radiation is visible to the eye

The range of wavelengths for visible light is approximately from 380 nm to 750 nm. Since the given wavelength of magnesium's radiation is 285.2 nm, which is outside the visible range, the radiation is not visible to the eye.
04

Calculate the energy of the radiation

To calculate the energy of the electromagnetic radiation, we can use the Planck's constant, \(h = 6.626 × 10^{-34} Js\), and the frequency, ν, in the following equation: \(E = hν\) Plug in the values of Planck's constant (h) and the frequency (ν) into the equation: \(E = (6.626 × 10^{-34} Js)(1.052 × 10^{15} Hz) \approx 6.97 × 10^{-19} J\) The energy of the electromagnetic radiation emitted by magnesium is approximately \(6.97 × 10^{-19} J\).

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Most popular questions from this chapter

Write the full ground-state electron configuration for \(\mathrm{Ca}^{2+}, \mathrm{S}^{2-}, \mathrm{Ar}, \mathrm{K}^{+}\)

Of the atoms \(\mathrm{Na}, \mathrm{Cl}, \mathrm{K}, \mathrm{Br}\), which has the largest atomic radius? Which has the largest first ionization energy?

How can you tell how many electrons a representative nonmetal is likely to gain? What, in general, will be the charge of the anion it forms?

Below are data on the first four ionization energies for a fictitious element \(X\). First ionization energy \(=500 \mathrm{~kJ} / \mathrm{mol}\) Second ionization energy \(=2000 \mathrm{~kJ} / \mathrm{mol}\) Third ionization energy \(=3500 \mathrm{~kJ} / \mathrm{mol}\) Fourth ionization energy \(=25,000 \mathrm{~kJ} / \mathrm{mol}\) From the data, which of the following statements is(are) incorrect? (a) \(\mathrm{X}\) could belong to group IIIA. (b) The fourth ionization energy is so much greater than the third ionization energy because \(\mathrm{X}^{3+}\) consists of a noble-gas core or a pseudo- noble-gas core. (c) The third ionization energy is greater than the second ionization energy because \(\mathrm{X}^{2+}\) has a bigger charge than \(\mathrm{X}^{+}\). (d) \(\mathrm{X}\) could belong to group IIIB. (e) \(X\) could belong to group VA.

According to the Bohr model, why do atoms get larger as you proceed down a group in the periodic table?
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