A near ultraviolet Photon of wavelength 300 nm is observed by a gas and then remitted as 2 photons.One Photon is red, wavelength of which is 760 nm. What would be the wavelength of second Photon in manometer?

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Solution

E =hc/λ

Wavelength of 1^st photon = λ₁

Wavelength of 2^nd photon = λ₂

E(Total) = E1 + E2 = Emitted energy

hc/ λ = hc/ λ₁ + hc/ λ₂

1/300 = 1/760 + 1/ λ₂

1/ λ₂ = (1/300 – 1/760)

λ₂ = 496 nm = wavelength of second photon

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A photon of wavelength 300 nm is absorbed by a gas and then reemitted as two photons. One photon is red with wavelength of 760 nm. The wave number of second photon will be

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A near ultraviolet Photon of wavelength 300 nm is observed by a gas and then remitted as 2 photons.One Photon is red, wavelength of which is 760 nm. What would be the wavelength of second Photon in manometer?

E =hc/λ Wavelength of 1st photon = λ1 Wavelength of 2nd photon = λ2 E(Total) = E1 + E2 = Emitted energy hc/ λ = hc/ λ1 + hc/ λ2 1/300 = 1/760 + 1/ λ2 1/ λ2 = (1/300 – 1/760) λ2 = 496 nm = wavelength of second photon

E =hc/λ

Wavelength of 1^st photon = λ₁

Wavelength of 2^nd photon = λ₂

E(Total) = E1 + E2 = Emitted energy

hc/ λ = hc/ λ₁ + hc/ λ₂

1/300 = 1/760 + 1/ λ₂

1/ λ₂ = (1/300 – 1/760)

λ₂ = 496 nm = wavelength of second photon