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Atom 0.65 Formula

Nov 26, 2017 An electron changes from an #n = 2# to an #n = 6# energy state. What is the energy of the photon in joules? Chemistry Bohr Model of the Atom Atoms and Electromagnetic Spectra. The Eureka Atom 65 is an espresso focused all-purpose home espresso grinder with 65mm flat steel burrs. This burr grinder uses a digital timed dosing system and stepless adjustment, as well as having a hands-free portafilter holder and a grounds bin.

1 Answer

Nov 27, 2017

Explanation:

The question wants you to determine the energy that the incoming photon must have in order to allow the electron that absorbs it to jump from #n_i = 2# to #n_f = 6#.

A good starting point here will be to calculate the energy of the photon emitted when the electron falls from #n_i = 6# to #n_f = 2# by using the Rydberg equation.

#1/(lamda) = R * (1/n_f^2 - 1/n_i^2)#

Here

#lamda# si the wavelength of the emittted photon
#R# is the Rydberg constant, equal to #1.097 * 10^(7)##'m'^(-1)#

Plug in your values to find

#1/lamda = 1.097 * 10^7color(white)(.)'m'^(-1) * (1/2^2 - 1/6^2)#

#1/lamda = 2.4378 * 10^6 color(white)(.)'m'^(-1)#

This means that you have

#lamda = 4.10 * 10^(-7)color(white)(.)'m'#

So, you know that when an electron falls from #n_i = 6# to #n_f = 2#, a photon of wavelength #'410 nm'# is emitted. This implies that in order for the electron to jump from #n_i = 2# to #n_f = 6#, it must absorb a photon of the same wavelength.

To find the energy of this photon, you can use the Planck - Einstein relation, which looks like this

#E = h * c/lamda#

Here

#E# is the energy of the photon
#h# is Planck's constant, equal to #6.626 * 10^(-34)color(white)(.)'J s'#
#c# is the speed of light in a vacuum, usually given as #3 * 10^8 color(white)(.)'m s'^(-1)#

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As you can see, this equation shows you that the energy of the photon is inversely proportional to its wavelength, which, of course, implies that it is directly proportional to its frequency.

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Plug in the wavelength of the photon in meters to find its energy

#E = 6.626 * 10^(-34) color(white)(.)'J' color(red)(cancel(color(black)('s'))) * (3 * 10^8 color(red)(cancel(color(black)('m'))) color(red)(cancel(color(black)('s'^(-1)))))/(4.10 * 10^(-7) color(red)(cancel(color(black)('m'))))#

#color(darkgreen)(ul(color(black)(E = 4.85 * 10^(-19)color(white)(.)'J')))#

I'll leave the answer rounded to three sig figs.

So, you can say that in a hydrogen atom, an electron located on #n_i = 2# that absorbs a photon of energy #4.85 * 10^(-19)##'J'# can make the jump to #n_f = 6#.

Atom 0.65 Vs

Atm065-a181

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