Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:
<span>E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| </span>
<span>E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J </span>
<span>After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:</span>
<span>E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m </span>
so 6.56×10^-7 m or better written 656 nm is in the visible spectrum
Answer:
22/0.50=44
Explanation:
average speed= distance / time
Q before connected = Q after connected C1V1+C2V2 = (C1+C2) V
C1= 3×10^-6 F
V1= 480v
C2= 4×10^-6 F
V2= 500v
(3×10^-6)×(480) + (4×10^-6)×(500) = (3×10^-6 + 4×10^-6) × V
Simplifying the above, we get:
( 1440× 10^-6) + (2000 ×10^-6) = (7 × 10^-6) × V.
Further simplified as:
3440 × 10^-6 = 7 × 10^-6 × V
Making V the subject
V = 491.43volts
Therefore the potential difference across each capacitor is 491.43v
