. Suppose an infrared photon with a wavelength of 853 nm were combined with another infrared photon with a wavelength of 865 nm, what would be the wavelength of the new 'combined' photon, in nm

Explanation:

Given the data in the question;

wavelength λ₁ = 853 nm = 853 × 10⁻⁹ m

wavelength λ₂ = 865 nm = 865 × 10⁻⁹ m

we determine the energy of each photons using the following expression;

= hc / λ

where h is Planck's constant ( 6.626 × 10⁻³⁴ J.s )

c is speed of light ( 3 × 10⁸ m/s )

For Photon with wavelength λ₁ = 853 × 10⁻⁹ m

= hc / λ

we substitute

= ( ( 6.626 × 10⁻³⁴ J.s )( 3 × 10⁸ m/s ) ) / (853 × 10⁻⁹ m)

= 1.9878 × 10⁻²⁵ / 853 × 10⁻⁹

= 2.33 × 10⁻¹⁹ J

For Photon with wavelength  λ₂ = 865 × 10⁻⁹ m

= ( ( 6.626 × 10⁻³⁴ J.s )( 3 × 10⁸ m/s ) ) / (865 × 10⁻⁹ m)

= 1.9878 × 10⁻²⁵ / 865 × 10⁻⁹

= 2.298 × 10⁻¹⁹ J

We know that; energy of the combined photon will be equal to the sum of energies of the two photons.

so

Energy of the combined photon = +

= 2.33 × 10⁻¹⁹ J + 2.298 × 10⁻¹⁹ J

 = 4.628 × 10⁻¹⁹ J

so wavelength of the new combined photon will be;

= hc / λ

λ = hc

λ = hc /

we substitute

λ = ( ( 6.626 × 10⁻³⁴ J.s )( 3 × 10⁸ m/s ) ) / 4.628 × 10⁻¹⁹ J

= 1.9878 × 10⁻²⁵ / 4.628 × 10⁻¹⁹

= 4.2952 × 10⁻⁷ m

= ( 4.2952 × 10⁻⁷ × 10⁹ )nm

= 429.5 nm

Therefore, Wavelength of the new 'combined' photon is 429.5 nm

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a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.