Question

1. The lines observed in H-lamp are parts of Balmer Series. All these lines are due to a transition from various higher n value to a common n. What is this common lower n value for all these lines?

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2. The lines on Absorption Atomic Spectra correspond to energies needed for electrons to be excited from a lower energy level to a higher energy level. Assume that the energy needed for an electron in 2p orbital in an O atom to jump to 3s orbital is 3.6*10-19 J, what is its wavelength of the line atomic spectra in nanometer (nm)?

3. Given: In Atomic Spectra lab, a student obtained his best-fit line equation to be y = 0.29 x + 46.8 when he plotted his Vernier reading on the y-axis and wavelength in nm on the x-axis.

4. The only electron in a hydrogen atom moved from n=2 to n=6. What is the wavelength of photon (in nm) being associated with this transition? Use whole numbers of nm without any decimal places, such as "415", "1125".

5.

Which of the following transitions in H will absorb the longest wavelength photon?

n=1 to n=2

n=2 to n=4

n=10 to n=3

n=3 to n=100

6. A student obtained the Rydberg constant to be 2.17*10-18 J. What is the % error of the measurement? Answer as percentage and keep one decimal place.

Answer 1

1.

Balmer series uses the 2nd electron transition, therefore n = 2

2. The lines on Absorption Atomic Spectra correspond to energies needed for electrons to be excited from a lower energy level to a higher energy level. Assume that the energy needed for an electron in 2p orbital in an O atom to jump to 3s orbital is 3.6*10-19 J, what is its wavelength of the line atomic spectra in nanometer (nm)?

E = h*c/WL

h = planck constant = 6.636*10^-34

c = sepeed of light = 3*10^8

WL = hc/E

WL = (6.36*10^-34)(3*10^8)/(3.6*10^-19) = 5.3*10^-7

WL = in nm multiply by 10^9 then 530 nm

3.

question?

4.

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)

E = (-2.178*10^-18)*(1/6^2 – 1/2 ^2)

E = 4.84*10^-19 J

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 = (6.626*10^-34)(3*10^8)/(4.84*10^-19) = 4.107*10^-7 m

in nm, 10^9 so

4.107*10^-78 = 410 nm

NOTE:

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