Possible sources of error in emission spectra of mercury?
Misreading the vernier scale. Not seeing the maximum left angle or right angle spectra, thus the average angle calculation doesn't match with the right wavelength for mercury. Difficulty in seeing the lines through the telescope.
Possible sources of error when performing absorption spectroscopy and wavelengths ?
making the solid sample too thick which will cause the pellet to be non-transparent and that will give you wide peaks which may become a problem when you try to interpret the graph. Also if insufficient mixture was originally added, the pellet will not remain on the grooves of the die. as well as having a wet solid mixture, the pellet will collapse upon removal of the bolts. In either case, the procedure needs to be repeated to avoid a poor/inaccurate graphs. hope that helps :)
What are possible sources of errors in a diffraction grating experiment when I get wavelengths which are 50% larger than they should be?
In order of decreasing liklihood from my experience.Are you measuring the separation of the maxima form the central (n=0) position. It is common to measure n=1 to n=1 on either side to reduce uncertainty but one has to remember to divide by 2.Are you using line spacing for your diffraction grating ie is it really what you think eg 80 lines /mm?An approximate error of 2 could be introduced if you think the grating is n lines/inch instead of n lines /cm or the other way round.Are you measuring the angle theta by measuring D (distance grating to screen) and x- separtation of maxima. Depending on apparatus used I have seen students measure light source to screen rather than grating to screen.
If I am testing the same solutions with a titration and a spectrometer, would the spectrometer show to be more?
there is a really accurate equation that goes with spectrometry; its called beer's law: A=abc Molar absorbivity=wavelength constant * cell length * concentration. A spectrometer would be more accurate, because you don't really have any limitations like titrations do. IN titrations you need to now the concentration of the titrant and therefore is a limitation. Also, there is much room for error in titrations, while a spectrometer has one source of error: getting contaminants on the surface of the cell.
Why is mercury light used in a diffraction grating experiment?
reference: http://people.uncw.edu/olszewski...Mercury is a heavy element (Atomic Number = 80) that exists as a liquid at room temperature, and is easily volatilized to a vapor. This makes it an ideal candidate for using in a mercury lamp.Mercury atoms, when excited by electric current, can emit a number of different frequencies (wave lengths) because there are a number of Hg atomic orbitals that are accessible for electronic transitions. The energy that is emitted during de-excitation of electrons from a number of these excited states are in the visible range of the electromagnetic spectrum. These lines can then be used as a source for diffraction grating studies, particularly to study optical interference phenomenon.In contrast, a sodium vapor lamp emits only a single yellow wavelength (D-line) in the visible range of the spectrum. Also, if you used an incandescent lamp, there is a plethora of visible lines that are emitted, which would confound and complicate the diffraction grating experiment.In theory, other elements can also be used in place of mercury, but the easy availability and high volatility of mercury made it a fairly obvious choice to develop a vapor lamp, as a source of visible light of specific frequencies.
What happens when a monochromatic light is replaced by a source of white light in a Young double-slit experiment?
White light consists of waves of inumerable wavelengths starting from violet to red colour. Therefore if monochromatic light in Young’s interference experiment is replaced by white light, then the waves of each wavelength form their separate interference patterns. The resultant effect of all these patterns is obtained on the screen.The path difference between waves starting from S1 and S2at the location (M) of central fringe is zero, i.e., for point M of screen S1m− S2m=0i.e., the waves of all colours reach at mid point M in same phase. Therefore the central fringe (at M) is white. As fringe width , i.e., wavelength increases in order of colours denoted by VIBGYOR therefore on either side of it some coloured fringes are obtained in order of colour VIBGYOR. That is the violet (V) fringe appears first and the red (R) the last. After this the fringes of many colours overlap at each point of the screen and so the screen appears uniformly illuminated.Thus if we use white light in place of monochromatic light the central fringe is white, containing on either side a few coloured fringes (in order VIBGYOR) and the remaining screen appears uniformly illuminated.Source:=>Xam idea · Physics Use Of White Light In Double Slit Experiment And Diffraction
What is the energy per mole of photons emitted?
The energy (E) per photon can be found from Planck's formula, E = hf, where f is the frequency and h is Planck's constant, equal to 6.625 * 10**-34 J-sec. We have the wavelength (w), which is related to the frequency by w = c/f, where c is the speed of light, equal to 3.0 * 10**8 meters per sec. Then f = c/w, so we can find the energy per photon: E = hc/w Since h & c are constant, hc is constant, equal to 19.875 * 10**-26 J-meters. So E = hc/w = (19.875 * 10**-26 J-m) / (5.147 * 10**-7 m.) Do the division and E = 3.8614 * 10**-19 joule per photon. To find the energy in a mole of these photons, we multiply the energy per photon by Avogadro's number, 6.023 * 10**23 and we get 2.325 * 10**5 joules, or 232.5 kilojoules per mole of photons.
How can you calculate the predicted wavelengths for the spectral transitions of the hydrogen?
Spectral transition means that an electron jumps from a higher shell in the atom to a lower shell. By this transition the electron changes from higher energy level to a lower one. The energy difference is emitted as light photon. Since there are only distinct energy levels, specified by the (integer) principal quantum number n, there are distinct spectral lines representing each transition. You find the wavelength from the Rydberg formula: 1/λ = RH · ( 1/(n1)² - 1/(n2)² ) λ is wavelength of the light emitted by transitions from n2 to n1. RH=1.09677583×10^7 m^-1 is the Rydberg constant for hydrogen. You can check your results by comparing with reported values of the Balmer series (3rd link). This series contains the spectral lines of all n→ 2 transitions.