For crystal diffraction experiments (discussed in Section 39.139.1), wavelengths on the order of 0.200.20 nm are often appropriate. Find the energy in electron volts for a particle with this wavelength if the particle is a photon.

Ch 39: Particles Behaving as Waves

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 39: Particles Behaving as Waves

Problema 2a

Capitolo 39, Problema 2a

For crystal diffraction experiments (discussed in Section $39.1$ ), wavelengths on the order of $0.20$ nm are often appropriate. Find the energy in electron volts for a particle with this wavelength if the particle is a photon.

Guida verificata passo dopo passo

Step 1: Recall the relationship between the energy of a photon and its wavelength, which is given by the equation:

E = rac{hc}{\(\text{λ}\)}

, where

h

is Planck's constant (

6.626 	imes 10^{-34} \(\text{ J·s}\)

c

is the speed of light (

3.00 	imes 10^8 \(\text{ m/s}\)

), and

\(\text{λ}\)

is the wavelength in meters.

Step 2: Convert the given wavelength from nanometers to meters. Since

1 \(\text{ nm}\) = 10^{-9} \(\text{ m}\)

, the wavelength

\(\text{λ}\) = 0.20 \(\text{ nm}\)

becomes

\(\text{λ}\) = 0.20 	imes 10^{-9} \(\text{ m}\)

Step 3: Substitute the values of

h

c

, and

\(\text{λ}\)

into the energy equation:

E = rac{(6.626 	imes 10^{-34})(3.00 	imes 10^8)}{0.20 	imes 10^{-9}}

. Perform the calculation to find the energy in joules.

Step 4: Convert the energy from joules to electron volts (eV). Use the conversion factor

1 \(\text{ eV}\) = 1.602 	imes 10^{-19} \(\text{ J}\)

. Divide the energy in joules by this factor to obtain the energy in electron volts.

Step 5: The final result will give the energy of the photon in electron volts. Ensure the units are consistent throughout the calculation and verify the result for accuracy.

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