Question

Image that the radiation emitted by the nitrogen at a frequency of 8.88×1014 Hz is absorbed by an electron in a molecule of methyl salicylate. As a result, the electron in the wintergreen oil molecule jumps to an excited state. Before returning to its ground state, the electron drops to an intermediate energy level, releasing two-thirds of the energy previously absorbed and emitting a photon. What is the wavelength of the photon emitted by the wintergreen oil molecule?

Answer 1

The wavelength (λ) of the photon : 5.106.10⁻⁷ m

Further explanation

The photoelectric effect is an electron coming out of a metal because of electromagnetic radiation

One type of electromagnetic radiation is light

Electrons can come out of metal because they absorb electromagnetic energy radiated on metals. There is also kinetic energy released from metal, which is according to the equation:

$\large {\boxed {\bold {E = hf-hfo}}}$

fo = the threshold frequency of electromagnetic waves

Radiation energy is absorbed by photons

The energy in one photon can be formulated as

$\rm E = h \times f$

Where

E = energy of light, J

h = Planck's constant (6,626.10⁻³⁴ Js)

f = Frequency of electromagnetic waves, Hz

f = c / λ

c = speed of light

= 3.10⁸

λ = wavelength

The radiation emitted by the nitrogen and absorbed by an electron

E = h. f

E = 6,626.10⁻³⁴. 8.88 × 10¹⁴

E = 5.839.10⁻¹⁹ J

Two-thirds of the energy previously absorbed and emitting a photon

E photon = 2/3 x E electron

E photon = 2/3 x 5.839.10⁻¹⁹ J

E photon = 3.893.10⁻¹⁹ J

$\rm E=\dfrac{hc}{\lambda}\\\\\lambda =\dfrac{hc}{E}\\\\\lambda=\dfrac{6.626\times 10^{-34}\times 3.10^8}{3,893.10^{-19}}\\\\\lambda=5.106.10^{-7}\:m$

Learn more  

Electromagnetic radiation with wavelength of 745 nm  

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The energy of a photon  

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The equation E = hf  

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the approximate energy of a photon  

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Answer 2

Answer:

$5.07\cdot 10^{-7} m$ (507 nm)

Explanation:

First of all let's calculate the energy of the photon absorbed by the electron, This is given by

$E=hf$

where

h is the Planck constant

$f=8.88\cdot 10^{14} Hz$ is the frequency of the photon

Substituting,

$E=(6.63\cdot 10^{-34}Js)(8.88\cdot 10^{14}Hz)=5.89\cdot 10^{-19} J$

The energy of the second photon, the one emitted when the electron drops to the intermediate energy level, is 2/3 of this energy:

$E'=\frac{2}{3}E=\frac{2}{3}(5.89\cdot 10^{-19} J)=3.92\cdot 10^{-19} J$

The relationship between the energy of the photon and its wavelength $\lambda$ is

$E=\frac{hc}{\lambda}$

where c is the speed of light. Solving for $\lambda$, we find the wavelength:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.92\cdot 10^{-19} J}=5.07\cdot 10^{-7} m$