Answer:
= 4
Explanation:
To solve this exercise we will use Bohr's atomic model
= - 13.606 / n² [eV]
The transition from level n = 2 to level n = 1 is valid
= - 13.606 [¼ -1/1]
= 10.2045 eV
Bohr's model for atoms with only one electron is
= -13.606 Z² / n²
Where Z is the atomic number of the atom.
In this case the helium atom has an atomic number of Z = 2 from the level n₀ = 2 let's look up to what level it reaches
ΔE = -13.606 [4 / ² - 4/4]
4 / ² = -ΔE / 13.606 + 1
4 / ² = -10.2045 / 13.606 +1 = -0.75 +1
4 / ² = 0.25
= √ 4 / 0.25
= 4
<span>A design is remodeled after analysis and tested again.</span>
Correct question is;
A thermal tap used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at 0°C.When the temperature is raised ,the fits is no longer exact. Calculate what change in temperature is necessary to produce a channel whose cross-sectional is equal to that of the tube of 1mm. (linear expansivity of silica = 8 × 10^(-6) /K and linear expansivity of aluminium = 26 × 10^(-6) /K).
Answer:
ΔT = 268.67K
Explanation:
We are given;
d1 = 8mm
d2 = 1mm
At standard temperature and pressure conditions, the temperature is 273K.
Thus; Initial temperature; T1 = 273K,
Using the combined gas law, we have;
P1×V1/T1 = P2×V2/T2
The pressure is constant and so P1 = P2. They will cancel out in the combined gas law to give:
V1/T1 = V2/T2
Now, volume of the tube is given by the formula;V = Area × height = Ah
Thus;
V1 = (πd1²/4)h
V2 = (π(d2)²/4)h
Thus;
(πd1²/4)h/T1 = (π(d2)²/4)h/T2
π, h and 4 will cancel out to give;
d1²/T1 = (d2)²/T2
T2 = ((d2)² × T1)/d1²
T2 = (1² × T1)/8²
T2 = 273/64
T2 = 4.23K
Therefore, Change in temperature is; ΔT = T2 - T1
ΔT = 273 - 4.23
ΔT = 268.67K
Thus, the temperature decreased to 268.67K
A) experiment. Is the answer.
hypothesis is the educated guess about what the result of the experiment is before conducting the experiment.
Observation is what you see and record during the experiment.
Can you put the answers, if there are any?
Once you do, Ill respond with a answer asap!
:)
