The element in a fluorescent lightbulb that absorbs UV light and releases visible light energy is ____?
2 answers:
You might be interested in
We use the criterion of significant figures to find the result with reliable figures
X = 9.2 10-5
now with the propagation of errors we obtain the result with its uncertainty
X ± ΔX = (9.2 ± 0.5) 10⁻⁵
given Parameter
* expression values with their absolute errors
to find
* the result with the correct significant figures
* the absolute error of the expression
Significant figures are defined with the number of decimals that give information, the number of figures in a quantity gives information about the uncertainty of this quantity.
There are two criteria for applying significant figures:
* Add and subtract the result of going with the number of decimal places of the figure that has the least
* Product and division as a result of going with the least number of significant figures than the value that has the least.
Remember that the zero to the left do not form a pair of the significant figures
Let's apply this belief to the case presented, let's write the precaution
where in this case they are worth
a = 0.0336 ± 0.0002
b = 0.010 ± 0.001
c = 255.4 ± 0.4
We see that the significant figures of each parameterize (a, b, c) and their absolute errors are correct.
Let's apply the criteria to the operation
a-b = 0.0336 - 0.010
a- b = 0.0236
we apply the criterion of significant figures for the subtraction, the result must be with 3 decimal places
a - b = 0.024
let's do the other operation
X =
X = 0.024 / 255.4
X = 9.24 10⁻⁵
We apply the criterion of significant figures for the division, in this case the result is left with two significant figures
X = 9.2 10⁻⁵
The uncertainty or error of the measurements is of most importance as it determines how many significative figures are reliable at a given magnitude.
If the magnitudes are measured with some type of instrument, the absolute error is given by the appreciation of the instrument, if the magnitude is calculated using some equation, the errors must be propagated using the variations of each parameter in the worst case.
the uncertainty of the calculated quantity (X) is
let's perform the derivatives
we substitute
remember that the bulk value guarantees that we tune the worst case. So all the mistakes add up
ΔX = Δa +
Δb +
Δc
ΔX = (Δa + Δb) +
Δc
we substitute
ΔX = (0.0002 + 0.001) +
0.4
ΔX = 4.698 10⁻⁶ + 1.45 10⁻⁷
DX = 4.8 10-6
Absolute errors must be given with a single significant figure
ΔX = 5 10⁻⁶
The result of the requested quantity using the criterion of significant figures and propagation of errors is
X ± ΔX = (9.2 ± 0.5) 10⁻⁵
learn more about significative figure here: