Answer:
1.082 mm
Explanation:
From the question, we can see that we were given The following
Wavelength of the atoms, λ = 502 nm = 502*10^-9 m
Radius of the screen away from the double slit, r = 1.1 m
We know that Y(20) = 10.2 mm = 10.2*10^-3 m
d = (20 * R * λ) / Y(20)
d = (20 * 1.1 * 502*10^-9)/10.2*10^-3
d = 1.1*10^-5 / 10.2*10^-3
d = 1.082 mm
Therefore, we can say that the distance of separation between the two slits is 1.082 mm
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>
The planet would stay in the same orbit but start revolving faster.
(Its year would get shorter.)
Answer:
Explanation:
As per the equation of voltage on capacitor we know that
now we know that voltage reached to its 80% of maximum value in 4 second time
so we will have
as we know that
