One of the primary visible emissions from a distant planet occurs at 425 nm. Calculate the energy of a mole of photons of this emission.]

Respuesta :

Answer:

Explanation:

Given

Wavelength of incoming light [tex]\lambda =425\ nm[/tex]

We know

[tex]speed\ of\ wave=frequency\times wavelength[/tex]

[tex]frequency=\frac{speed}{wavelength}[/tex]

[tex]\mu =\frac{3\times 10^8}{425\times 10^{-9}}[/tex]

[tex]\mu =7.058\times 10^{14}\ Hz[/tex]

Energy associated with this frequency

[tex]E=h\mu [/tex]

where h=Planck's constant

[tex]E=6.626\times 10^{-34}\times 7.058\times 10^{14}[/tex]

[tex]E=46.76\times 10^{-20}\ Hz[/tex]

Energy of one mole of Photon[tex]=N_a\times E[/tex]

[tex]=6.022\times 10^{23}\times 46.76\times 10^{-20}[/tex]

[tex]=281.58\times 10^{3}[/tex]

[tex]=281.58\ kJ[/tex]

The Energy of a single photon of 425 nm emission is 4.677 × 10⁻¹⁹J,

While the energy of a mole of photons of the same emission is 281.6kJ

Given the data in the question;

Wavelength; [tex]\lambda = 425nm = 4.25*10^{-7}m[/tex]

Energy of a mole of photons of this emission; [tex]E_{1mole} = \ ?[/tex]

First we determine the Frequency, from the relation between wavelength, frequency and the velocity or speed of light:  

[tex]f = \frac{c}{\lambda}[/tex]

Where f is the frequency , λ is the wavelength and c is the speed of light( [tex]3*10^8m/s[/tex])

We substitute in our values

[tex]f = \frac{3*10^8m/s}{4.25*10^{-7}m} \\\\f = 7.0588*10^{14}Hz[/tex]

Now, We find the Photon energy:

[tex]E = hf[/tex]

Where:

E is the photon energy,

h is the Planck's constants ( [tex]6.626*10^{-34}JHz^{-1}[/tex] ) and

f is the frequency.

So we substitute in our values

[tex]E = ( 6.626*10^{-34}JHz^{-1}) * ( 7.0588*10^{14}Hz\\\\E = 4.677*10^{-19}J[/tex]

Now, we know that, Energy of one mole of Photon = [tex]N_A * E[/tex]

Where E is the photon energy and [tex]N_A[/tex] is the Avogadro number ( [tex]6.022 * 10^{23}[/tex])

So,

Energy of a mole of photons of this emission:

[tex]E_{1mole} = (6.022*10^{23}) * ( 4.677*10^{-19}J)\\\\E_{1mole} = 281648.94 J\\\\E_{1mole} = 281.6 kJ[/tex]

Therefore, the Energy of a single photon of 425 nm emission is 4.677 × 10⁻¹⁹J,

While the energy of a mole of photons of the same emission is 281.6kJ

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