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Chapter 17: Problem 2416

Calculate the energy of a photon of radian wavelength \(6000 \AA\) in \(\mathrm{eV}\) (A) \(20.6 \mathrm{eV}\) (B) \(2.06 \mathrm{eV}\) (C) \(1.03 \mathrm{eV}\) (D) \(4.12 \mathrm{eV}\)

Short Answer

Expert verified
The energy of a photon with a wavelength of \(6000 \mathrm{\AA}\) is approximately \(2.07 \mathrm{eV}\), which is closest to option (B), \(2.06 \mathrm{eV}\).

Step by step solution

01

Write down the formula for the energy of a photon

The formula for the energy of a photon is given by: \(E = \dfrac{hc}{\lambda}\), where \(E\) is the energy of the photon, \(h\) is Planck's constant (approximated to \(6.63 \times 10^{-34} \mathrm{J·s}\)), \(c\) is the speed of light (approximated to \(3 \times 10^8 \mathrm{m/s}\)), and \(\lambda\) is the wavelength of the photon.
02

Convert the given wavelength from angstroms to meters

The wavelength is given as \(6000 \mathrm{\AA}\) (angstroms). To convert this to meters, we can use the conversion factor \(1 \mathrm{\AA} = 10^{-10} \mathrm{m}\). \(\lambda = 6000 \mathrm{\AA} \times \dfrac{10^{-10} \mathrm{m}}{1 \mathrm{\AA}} = 6000 \times 10^{-10} \mathrm{m}\)
03

Calculate the energy of the photon in joules

Using the formula from Step 1 and the wavelength in meters, we can calculate the energy of the photon in joules: \(E = \dfrac{(6.63 × 10^{-34} \mathrm{J·s}) \times (3 × 10^8 \mathrm{m/s})}{(6000 × 10^{-10} \mathrm{m})}\) \(E = \dfrac{1.989 × 10^{-25} \mathrm{J·m}}{6 × 10^{-7} \mathrm{m}}\) \(E = 3.315 × 10^{-19} \mathrm{J}\)
04

Convert energy from joules to electron volts

To convert the energy from joules to electron volts, we can use the conversion factor \(1 \mathrm{eV} = 1.602 \times 10^{-19} \mathrm{J}\): \(E_\mathrm{eV} = \dfrac{3.315 \times 10^{-19} \mathrm{J}}{1.602 \times 10^{-19} \mathrm{J/eV}}\) \(E_\mathrm{eV} \approx 2.07 \mathrm{eV}\)
05

Determine the correct answer from the options given

Based on our calculations, the correct answer is approximately \(2.07 \mathrm{eV}\) which is closest to option (B). So the correct answer is: (B) \(2.06 \mathrm{eV}\)

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Most popular questions from this chapter

Uncertainty of momentum of particle is $10^{-3} \mathrm{~kg} \mathrm{~ms}^{-1}\( so minimum uncertainty in its position is \)\ldots \ldots \mathrm{m}$. (A) \(10^{-8} \mathrm{~m}\) (B) \(10^{-12} \mathrm{~m}\) (C) \(10^{-16} \mathrm{~m}\) (D) \(10^{-4} \mathrm{~m}\)

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