A near UV photon of $300 n m$ is absorbed by a gas and then re-emitted as two photons. One photon is red light with wavelength $760 n m$ , then the wavelength of the second photon is:

1060 n m

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496 n m

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300 n m

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215 n m

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Solution

The correct option is B $496 n m$
Since, energy values are additives.

$E = E_{1} + E_{2}$

$\frac{h c}{λ} = \frac{h c}{λ_{1}} + \frac{h c}{λ_{2}}$

$\frac{1}{300} = \frac{1}{760} + \frac{1}{λ_{2}}$

$\frac{1}{λ_{2}} = \frac{1}{300} - \frac{1}{760}$

$λ = 495.6 n m = 496 n m$

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Similar questions

Q. A photon of

300 n m

is absorbed by gas and then re-emits two photons. One re-emitted photon has wavelength

496 n m

, the wavelength of the second re-emitted photon is:

Q. A photon of

300 n m

is absorbed by a gas which then re-emits two photons. One re-emitted photon has wavelength

496 n m

. Calculate the energy associated with the other photon re-emitted out.

Q. A near ultraviolet photon of

300 n m

is absorbed by a gas and then remitted as two photons. One photon is red with wavelength of

760 n m

. What would be the wave number of the second photon?

A near UV photon of 300 nm is absorbed by a gas and then re-emitted as 2 photons. One photon is red with wavelength 760 nm. Hence wavelength of the second photon is:

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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A near UV photon of 300 nm is absorbed by a gas and then re-emitted as two photons. One photon is red light with wavelength 760 nm, then the wavelength of the second photon is:

The correct option is B 496 nmSince, energy values are additives.E=E1+E2hcλ=hcλ1+hcλ21300=1760+1λ21λ2=1300−1760λ=495.6 nm=496 nm

A near UV photon of $300 n m$ is absorbed by a gas and then re-emitted as two photons. One photon is red light with wavelength $760 n m$ , then the wavelength of the second photon is:

The correct option is B $496 n m$
Since, energy values are additives.

$E = E_{1} + E_{2}$

$\frac{h c}{λ} = \frac{h c}{λ_{1}} + \frac{h c}{λ_{2}}$

$\frac{1}{300} = \frac{1}{760} + \frac{1}{λ_{2}}$

$\frac{1}{λ_{2}} = \frac{1}{300} - \frac{1}{760}$

$λ = 495.6 n m = 496 n m$