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A photon of ultraviolet (UV) light possesses enough energy to mutate a strand of human DNA. What is the energy of a single UV photon and a mole of UV photons having a wavelength of 25 nm?

Short Answer

Expert verified
The energy of a single UV photon with a wavelength of 25 nm is \(E = \frac{6.626 \times 10^{-34} Js \cdot 2.998 \times 10^8 m/s}{25 \times 10^{-9} m} = 7.97 \times 10^{-18} J\). To find the energy of a mole of UV photons, we multiply the energy of a single photon by Avogadro's number: \(Energy\_of\_mole\_of\_photons = E \times 6.022 \times 10^{23} photons/mole = 7.97 \times 10^{-18} J \times 6.022 \times 10^{23} photons/mole = 4.80 \times 10^6 J/mole\). Therefore, the energy of a single UV photon is \(7.97 \times 10^{-18} J\) and the energy of a mole of UV photons is \(4.80 \times 10^6 J/mole\).

Step by step solution

01

Calculate the energy of a single photon

To calculate the energy of a single photon, we'll use Planck's equation: E = h * c / λ Where: E = energy of a single photon, h = Planck's constant (6.626 × 10^-34 Js), c = speed of light (2.998 × 10^8 m/s), and λ = wavelength of the UV photons (25 nm which is equal to 25 x 10^-9 m) E = (6.626 × 10^-34 Js) * (2.998 × 10^8 m/s) / (25 × 10^-9 m)
02

Find the energy of a mole of photons

To find the energy of a mole of photons, we'll multiply the energy of a single photon by Avogadro's number: Energy of a mole of photons = E_single_photon × Avogadro's_number Let N_A = Avogadro's number (6.022 × 10^23 photons/mole) Energy_of_mole_of_photons = E × N_A Now, let's compute the energy of a single photon and a mole of photons.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Planck's Equation
To understand the energy of a single photon, we make use of Planck's equation. This fundamental equation allows us to calculate the energy of a photon based on its wavelength and is given by:\[ E = \frac{h \cdot c}{\lambda} \]Where:- \( E \) is the energy of the photon.- \( h \) is Planck's constant, valued at \( 6.626 \times 10^{-34} \text{Js} \), a fundamental constant in physics.- \( c \) is the speed of light, approximately \( 2.998 \times 10^8 \text{m/s} \).- \( \lambda \) is the wavelength of the radiation in meters.This equation tells us that as the wavelength \( \lambda \) decreases, the energy \( E \) of a photon increases. For ultraviolet light with a short wavelength of 25 nm, which is equivalent to \( 25 \times 10^{-9} \text{m} \), the energy is relatively high. This high energy is what makes UV photons capable of inducing mutations in DNA.
Avogadro's Number
Avogadro's number is a fundamental concept in chemistry that allows us to relate the small world of atoms and molecules to the macroscopic quantities we deal with in daily life. Avogadro's number is approximately \( 6.022 \times 10^{23} \) and represents the number of atoms, ions, or molecules in one mole of a substance.

Application in Photon Energy Calculation

When you want to calculate the energy of not just one photon, but a whole mole of photons, Avogadro's number becomes essential. By multiplying the energy of a single photon by Avogadro's number, you obtain the total energy of a mole of photons, which is especially useful in practical applications where handling individual photons isn't feasible.In the context of UV-light, knowing the energy per mole of photons helps in understanding how much energy is involved in processes like DNA mutations.
Wavelength Calculation
Understanding the wavelength of light is crucial for calculations involving photon energy. The wavelength \( \lambda \) is the distance over which the light wave's shape repeats, typically expressed in meters, nanometers, or angstroms.

Converting Wavelength Units

In scientific calculations, it's essential to convert units consistently. For UV light, you might be given the wavelength in nanometers (nm), like 25 nm. Since Planck's equation requires meters, convert this by recognizing that 1 nm equals \( 10^{-9} \text{m} \). Thus, 25 nm becomes \( 25 \times 10^{-9} \text{m} \).Knowing how to switch between different units of measurement ensures accuracy in energy calculations and helps unlock a deeper understanding of how light's properties affect its interactions with matter.

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Most popular questions from this chapter

Neutron diffraction is used in determining the structures of molecules. a. Calculate the de Broglie wavelength of a neutron moving at 1.00\(\%\) of the speed of light. b. Calculate the velocity of a neutron with a wavelength of 75 \(\mathrm{pm}\left(1 \mathrm{pm}=10^{-12} \mathrm{m}\right)\)

In defining the sizes of orbitals, why must we use an arbitrary value, such as 90% of the probability of finding an electron in that region?

Answer the following questions based on the given electron configurations, and identify the elements. a. Arrange these atoms in order of increasing size: \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{6} ;[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{1} ;[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{3}\) b. Arrange these atoms in order of decreasing first ionization energy: [Ne \(3 s^{2} 3 p^{5} ;[\operatorname{Ar}] 4 s^{2} 3 d^{10} 4 p^{3} ;[\operatorname{Ar}] 4 s^{2} 3 d^{10} 4 p^{5}\)

Consider the following ionization energies for aluminum: $$\begin{array}{c}{\operatorname{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} \quad I_{1}=580 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} \quad I_{2}=1815 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} \quad I_{3}=2740 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} \quad I_{4}=11,600 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ a. Account for the trend in the values of the ionization energies. b. Explain the large increase between \(I_{3}\) and \(I_{4}\)

The four most abundant elements by mass in the human body are oxygen, carbon, hydrogen, and nitrogen. These four elements make up about 96% of the human body. The next four most abundant elements are calcium, phosphorus, magnesium, and potassium. Write the expected ground-state electron configurations for these eight most abundant elements in the human body.

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