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tiny-mole [99]
3 years ago
13

Hydrogen line spectrum lies entirely within visible range

Physics
1 answer:
mart [117]3 years ago
6 0

No, that's silly.

You've got your Pfund series where electrons fall down to the 5th level,
your Brackett series where they fall to the 4th level, and your Paschen
series where they fall to the 3rd level.  All of those transitions ploop out
photons at Infrared wavelengths.

THEN next you get your Balmer series, where the electrons fall in
to the 2nd level.  Most of those are at visible wavelengths, but even
a few of the Balmer transitions are in the Ultraviolet.

And then there's the Lyman series, where electrons fall all the way
down to the #1 level.  Those are ALL in the ultraviolet. 
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4 0
2 years ago
When you stretch a spring 20 cm past its natural length, it exerts a force of 8
Nastasia [14]

Answer:

40 N/m

Explanation:

F = -kx (This is the Hooke's Law equation)

F is the force the spring exerts = 8 N

-k = spring constant

x = displacement (The distance stretched past it's natural length) = 20cm

x needs to be in meters, and 20 cm is = to 0.2 meters

Finally:

8N = -k (0.2m)

-k = 8N / 0.2 m

k = -40 N/m

6 0
3 years ago
Read 2 more answers
In middle ages people believed that abnormal behavior was cause by ?
Svetach [21]
C, they didn't know any better
8 0
3 years ago
Read 2 more answers
Which of the following formulas is used to solve for current flowing through an electrical circuit? A. current = voltage ÷ resis
Alex787 [66]
The answer to your question is a

8 0
3 years ago
Read 2 more answers
A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th
Stells [14]

Answer:

The relationship between the initial stored energy PE_{i} and the stored energy after the dielectric is inserted PE_{f} is:

c) PE_{f} =0.5\ PE_{i}

Explanation:

A parallel plate capacitor with C_{o} that is connected to a voltage source V_{o} holds a charge of Q_{o} =C_{o} V_{o}. Then we disconnect the voltage source and keep the charge Q_{o} constant . If we insert a dielectric of \kappa=2 between the plates while we keep the charge constant, we found that the potential decreases as:

                                                     V=\frac{V_{o}}{\kappa}

The capacitance is modified as:

                                              C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}

The stored energy without the dielectric is

                                               PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}

The stored energy after the dielectric is inserted is:

                                               PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}

If we replace in the above equation the values of V and C we get that

                                         PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})

                                                   PE_{f} =\frac{PE_{i}}{\kappa}

Finally

                                                  PE_{f} =0.5\ PE_{i}

                                               

                                     

5 0
3 years ago
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