Title: A hydrogen atom jumps from energy level 5 to energy level 2 and emits a photon. What are the frequency and wav Post by: smrrdn on Apr 6, 2013 A hydrogen atom jumps from energy level 5 to energy level 2 and emits a photon. What are the frequency and wavelength of the photon emitted?
Title: A hydrogen atom jumps from energy level 5 to energy level 2 and emits a photon. What are the frequency and wav Post by: Solarin on Apr 6, 2013 use the eqn:
f=R*{(1/n1)^2~(1/n2)^2}, R=Rydberg constant=109678/cm n=energy level number and then c=f? for the wavelength. Title: A hydrogen atom jumps from energy level 5 to energy level 2 and emits a photon. What are the frequency and wav Post by: micaht25 on Apr 6, 2013 the electron in 5th energy level have energy=-0.54 eV
the electron in 2nd energy level have energy= -3.4eV thus energy released when it jumps from 5th to 2nd state= -0.54-(-3.4) = 2.86eV and 1eV = 1.6 x 10^-19 joule 2.86eV= 4.58 x 10^-19 joule from planck's formula hv = energy radiated = 4.58 x 10^-19 v= 6.19 x 10^14 hertz and wavelength = 4.18 x 10^-7 meter for more details mail me Title: A hydrogen atom jumps from energy level 5 to energy level 2 and emits a photon. What are the frequency and wav Post by: ik214shizo on Apr 6, 2013 First, find the energy of each level:
R=1.0974 x 10^7 m^-1 h= Planck's constant= 6.626 x 10^-34 J*s c=speed of light=3.0 x 10^8 m/s Energy level=[ -(R)(h)(c)] / (level # squared) Energy level 2= [ -(1.0974 x 10^7)(6.626x 10^-34)(3.0 x 10^8)] / 4 = -5.45 x 10^-19 J Energy level 5 = [ -(1.0974 x 10^7)(6.626x 10^-34)(3.0 x 10^8)] / 25 = -8.72 x 10^-20 J Then take the absolute value of the difference of the two levels' energies. |(-5.45 x 10^-19 J) - (-8.72 x 10^-20 J)|= 4.578 x 10^-19 J Then use that to find the frequency using the eqtn Energy=(h)(frequency) So 4.578 x 10^-19 = (6.626 x 10^-34)(frequency) frequency= 6.909 x 10^14 And then use the frequency to solve for the wavelength where (wavelength)(frequency)=speed of light So then wavelegth= (3.0 x 10^8) / (6.909 x 10^14) Which gives us 4.34 x 10^-7 m. Then convert to nm, which gives you 434 nm as your wavelength. Title: A hydrogen atom jumps from energy level 5 to energy level 2 and emits a photon. What are the frequency and wav Post by: smrj07 on Apr 6, 2013 several people have shown you the steps to find the frequency and wavelength of the photon, one note that might be helpful depending on your course is that for the hydrogen atom, transitions into the second energy level have a special name, these are called Balmer lines and are the visible lines in the spectrum of hydrogen
transitions into the first energy level are called the Lyman lines and are of higher energy and are in the ultraviolet part of the spectrum transitions into the third and higher levels have less energy and are in the infrared, so the Balmer series are the only transitions that produce visible light |