Question

In: Chemistry

Calculate the wavelengths (in nm) of the first 5 transitions in the Balmer series using the...

Calculate the wavelengths (in nm) of the first 5 transitions in the Balmer series using the Rydberg equation. Convert the wavelengths calculated into frequencies.

Expert Solution

Balmer series --> n = 2 so

n3 --> 2

n4 --> 2

n5 --> 2

N6 --> 2

n7 --> 2

Apply Rydberg Formula

E = R*(1/nf^2 – 1/ni ^2)

R = -2.178*10^-18 J

Nf = final stage/level

Ni = initial stage/level

E = Energy per unit (i.e. J/photon)

For the wavelength:

WL = h c / E

h = Planck Constant = 6.626*10^-34 J s

c = speed of particle (i.e. light) = 3*10^8 m/s

E = energy per particle J/photon

WL = wavelength in meters

simplified:

WL = (6.626*10^-34 )(3*10^8) / ((2.178*10^-18) * (1/nf^2 – 1/ni ^2)) * 10^9

since nf = 2, then

WL = (6.626*10^-34 )(3*10^8) / ((2.178*10^-18) * (1/2^2 – 1/ni ^2)) * 10^9

n = 3,4,5,6,7,

WL = (6.626*10^-34 )(3*10^8) / ((2.178*10^-18) * (1/2^2 – 1/3^2)) * 10^9 = 657 nm

WL = (6.626*10^-34 )(3*10^8) / ((2.178*10^-18) * (1/2^2 – 1/4^2)) * 10^9 = 487 nm

WL = (6.626*10^-34 )(3*10^8) / ((2.178*10^-18) * (1/2^2 – 1/5^2)) * 10^9 = 434 nm

WL = (6.626*10^-34 )(3*10^8) / ((2.178*10^-18) * (1/2^2 – 1/6^2)) * 10^9 = 410 nm

WL = (6.626*10^-34 )(3*10^8) / ((2.178*10^-18) * (1/2^2 – 1/7^2)) * 10^9 = 398 nm

change to frequency

v(3) = c/WL*10^9 = (3*10^8)*(10^9)/(657 )= 4.56*10^14 Hz

v(4) = c/WL*10^9 = (3*10^8)*(10^9)/(487 )= 6.16*10^14 Hz

v(5) = c/WL*10^9 = (3*10^8)*(10^9)/(434 )= 6.91*10^14 Hz

v(6) = c/WL*10^9 = (3*10^8)*(10^9)/(410 )= 7.32*10^14 Hz

v() = c/WL*10^9 = (3*10^8)*(10^9)/(398)= 7.54*10^14 Hz

queen_honey_blossom answered 2 years ago

4. The Balmer Series The Balmer series is limited to wavelengths in the visible regions from...

4. The Balmer Series The Balmer series is limited to wavelengths in the visible regions from 250 nm to 700 nm. On the basis of your entries in the above table and the transitions on your energy level diagram what common characteristic do the lines in the Balmer series have? What would be the shortest and longest possible wavelengths for a line in the Balmer series?

The Balmer series for the hydrogen atom corresponds to electronic transitions that terminate in the state...

The Balmer series for the hydrogen atom corresponds to electronic transitions that terminate in the state with quantum number n = 2 as shown in the figure below. Consider the photon of longest wavelength corresponding to a transition shown in the figure. (a) Determine its energy (b) Determine its wavelength Consider the spectral line of shortest wavelength corresponding to a transition shown in the figure (c) Find its photon energy (d) Find its wavelength (e) What is the shortest possible wavelength in the Balmer series?

Explore the Balmer series of atomic transitions for a hydrogen atom. Assume the shortest wavelength possible...

Explore the Balmer series of atomic transitions for a hydrogen atom. Assume the shortest wavelength possible occurs when the atom is ionized or when n = ∞ . What is the longest and shortest wavelengths possible? What is the frequency, wavelength and energy of the longest wavelength? longest λ = ___________ nm frequency = _________ Hz energy = _________ eV Shortest λ = ___________ nm

The only visible spectral lines of hydrogen are the four Balmer series lines (transitions to n=2)....

The only visible spectral lines of hydrogen are the four Balmer series lines (transitions to n=2). We wish to cause hydrogen gas to glow with all its characteristic visible colors. a) To how high an energy level must the electrons be excited? b) Energy is absorbed in collisions with other particles. Assume that after absorbing energy in one collision, an electron jumps down through lower levels so rapidly so that it is in the ground state before the collision occurs....

he K series of the discrete x-ray spectrum of tungsten contains wavelengths of 0.018 5 nm,...

he K series of the discrete x-ray spectrum of tungsten contains wavelengths of 0.018 5 nm, 0.020 9 nm, and 0.021 5 nm. The K-shell ionization energy is 69.5 keV. (a) Determine the ionization energies of the L, M, and N shells. L shell ___keV M shell ___keV N shell ___keV

Calculate the wavelengths of the following objects (in nm): a muon (a subatomic particle with a...

Calculate the wavelengths of the following objects (in nm): a muon (a subatomic particle with a mass of 1.884 × 10–25 g) traveling at 320.0 m/s an electron (me = 9.10939 × 10–28 g) moving at 3.90 × 106 m/s in an electron microscope an 76.0 kg athlete running a "4-minute mile" (i.e. 4.00 min/mile) Earth (mass = 5.90 × 1027 g) moving through space at 3.10 × 104 m/s

D. The Balmer Series 1. When Balmer found his famous series for hydrogen in 1886, he...

D. The Balmer Series 1. When Balmer found his famous series for hydrogen in 1886, he was limited experimentally to wavelengths in the visible and near ultraviolet regions from 250 nm to 700 nm, so all the lines in his series lie in that region. a) Based on the entries in Table 2 and the transitions on your energy level diagram, what common characteristic do the lines in the Balmer series have? B) What would be the longest possible wavelength...

compute the following: (a) Calculate the wavelengths for the Paschen series for Hydrogen (b) If the...

compute the following: (a) Calculate the wavelengths for the Paschen series for Hydrogen (b) If the average speed of a Hydrogen electron is 4.0*10^6 m/s with a 0.95% uncertainty. What is the uncertainty of its position? ( Hint: the mass of an electron is 9.109*10^-31)

Calculate the frequencies and energies for these 4 wavelengths: 410 nm, 435nm, 487nm, 655nm

Atomic hydrogen exhibits a series of emission spectral lines that includes the following wavelengths: 1875 nm,...

Atomic hydrogen exhibits a series of emission spectral lines that includes the following wavelengths: 1875 nm, 1282 nm, and 1094 nm. (a) What is the name of the series of emission spectral lines? (b) What is the wavelength of the next line in the series?