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Chapter 4: Problem 2

What is the energy of red light with a wavelength of \(660.5 \mathrm{~nm}\) ?

Short Answer

Expert verified
The energy of the red light with a wavelength of 660.5 nm is approximately \(2.9987 \times 10^{-19}\) joules.

Step by step solution

01

Convert wavelength to meters

We are given the wavelength of the red light as 660.5 nm. We need to convert this value to meters in order to use it in our calculations. 1 nm = 10^-9 meters So, 660.5 nm = 660.5 x 10^-9 meters
02

Calculate the energy of the red light

Now, let's plug in the values into the formula for the energy of a photon: Energy (E) = \( \dfrac{hc}{λ} \) Where: h = 6.626 x 10^-34 J·s (Planck's constant) c = 2.998 x 10^8 m/s (speed of light) λ = 660.5 x 10^-9 m (wavelength in meters) Energy (E) = \( \dfrac{(6.626 \times 10^{-34} \text{J} \cdot \text{s})(2.998 \times 10^{8} \text{m}/\text{s})}{660.5 \times 10^{-9} \text{m}} \)
03

Simplify and find the energy

Now, we will simplify this equation and find the energy of the red light. Energy (E) ≈ \( \dfrac{(6.626 \times 10^{-34})(2.998 \times 10^{8})}{660.5 \times 10^{-9}} \) J Energy (E) ≈ 2.9987 x 10^-19 J The energy of the red light with a wavelength of 660.5 nm is approximately 2.9987 x 10^-19 joules.

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Most popular questions from this chapter

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.

Arrange calcium, strontium, arsenic, bromine, and chlorine in order of (a) increasing atomic size and (b) increasing first ionization energy.

Which subshell has the lowest energy? (a) \(4 s\) (b) \(3 p\) (c) \(2 p\) (d) \(3 s\) (e) \(2 s\)

Identify the period 2 element that is described by the ionization data below. \(\mathrm{M}(\mathrm{g}) \rightarrow \mathrm{M}^{+} 1 \mathrm{e}^{-} \quad \mathrm{IE}(1)=1.40 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{+}(\mathrm{g}) \rightarrow \mathrm{M}^{2+} 1 \mathrm{e}^{-} \mathrm{IE}(2)=2.86 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{2+}(\mathrm{g}) \rightarrow \mathrm{M}^{3+} 1 \mathrm{e}^{-} \mathrm{IE}(3)=4.58 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{3+}(\mathrm{g}) \rightarrow \mathrm{M}^{4+} 1 \mathrm{e}^{-} \mathrm{IE}(4)=7.48 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{4+}(\mathrm{g}) \rightarrow \mathrm{M}^{5+} 1 \mathrm{e}^{-} \mathrm{IE}(5)=9.44 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{5+}(\mathrm{g}) \rightarrow \mathrm{M}^{6+} 1 \mathrm{e}^{-} \mathrm{IE}(6)=5.33 \times 10^{4} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{6+}(\mathrm{g}) \rightarrow \mathrm{M}^{7+} 1 \mathrm{e}^{-} \mathrm{IE}(7)=6.44 \times 10^{4} \mathrm{~J} / \mathrm{mol}\)

Predict the formula of the compound that forms when sodium atoms react with sulfur atoms. Completely explain your reasoning.
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