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Chapter 38: Q5P (page 1181)

The meter was once defined as 1650763.73 wavelengths of the orange light emitted by a source containing krypton-86 atoms. What is the photon energy of that light?

Short Answer

Expert verified

The photon energy of the light is 2.047 eV.

Step by step solution

01

Describe the expression of the energy of the photon.

The energy Eof a photon of wavelength λis given by,

role="math" localid="1663127601614" E=hcλ……. (1)

Here, h is Planck’s constant, and c is the speed of light.

02

Determine the photon energy of that light

It is given that,

1m=1650763.73wavelengthsoforangelight.

The wavelength of the orange light is,

λ=11650763.73m

Substitute the below values in eq 1.

λ=11650763.73mh=6.626×10-34J.sc=3×108m/s

E=6.626×10-34J.s3×108m/s11650763.73m=32833690.5897×10-26J=32833690.5897×10-26J1eV1.6×10-19J=2.047eV

Therefore, the photon energy of the light is 2.047 eV.

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