Answer:
λ = 623.2 nm
Explanation:
We are given;
separation distance; d = 0.195 mm = 0.195 × 10^(-3) m
interference pattern distance; D = 4.85 m
Width of two adjacent bright interference; w = 1.55 cm = 1.55 × 10^(-2) m
Formula for fringe width is given as;
w = λD/d
Where λ is wavelength
Thus;
λ = dw/D
λ = (0.195 × 10^(-3) × 1.55 × 10^(-2))/4.85
λ = 0.0000006232 m
Converting to nm gives;
λ = 623.2 nm
<span>A differences in the warmth, or moisture level as well as neighbouring areas of pressure in </span>air cause air<span> to circulate. in the earth's atmosphere.</span>
What it looks to be that you found in A was the "initial"...b/c the question asks:
<span>"how much energy does the electron have 'initially' in the n=4 excited state?" </span>
<span>"final" would be where it 'finally' ends up at, ie. its last stop...as for this question...the 'ground state' as in its lowest energy level. </span>
The answer comes to: <span>−1.36×10^−19 J</span>
You use the same equation for the second part as for part a.
<span>just have to subract the 2 as in the only diff for part 2 is that you use 1squared rather than 4squared & subract "final -initial" & you should get -2.05*10^-18 as your answer. </span>
never seen one although i guess the elegant answer is D) none of the mentioned posibilities
Ca(s) + O₂(g)→ 2 CaO (s)
The sequence of coefficient that will balance the above equation is
2,1 ,2
Explanation
Coefficient is the number in front of the formulas.
When balancing the equation the number of atoms in reactant side must be equal to number of atoms in product side.
The sequence that balance the equation above is 2,1,2 therefore the balanced equation is as below.
2Ca (s) + O₂ → 2CaO (s)