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3 years ago

Hydrogen line spectrum lies entirely within visible range

1 answer:

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.

You might be interested in

A 44-turn rectangular coil with length ℓ = 17.0 cm and width w = 8.10 cm is in a region with its axis initially aligned to a hor

Mumz [18]

Answer:

The maximum induced emf in the rotating coil = 29.66V

The induced emf in the rotating coil when (t = 1.00 s) = 26.66V

The maximum rate of change of the magnetic flux through the rotating coil = 0.674Wb/s

Explanation:

Lets state the parameters we are being given right from the question:

Number of rectangular coil, (N) = 44

Length of Coil, l =17cm in meters we have; (l) = 17 × 10⁻² m

Width of Coil, w =8.10cm in meters we have; (w) = 8.10 × 10⁻² m

Magnitude of Uniform Magnetic Field (B) = 767mT= 765 × 10⁻³ T

Angular Speed of Coil, (ω) = 64 rad/s

(a)

To calculate the induced emf in the rotating cell,we can use the formula:

emf = NBAωsin(ωt)

For maximum induced emf, the value of sin(ωt) will be 1

$emf_max$ = NBAω ; if (A = l × w) , we have:

$emf_max$ = NB(l × w)ω

subsitituting the parameters into the above equation; we have:

$emf_max$ = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64

= 29.66V

(b)

At t = 1s, the induced emf is calculated as:

emf = NBAωsin(ωt)

substituting the parameters into the equation, we have:

emf = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64 × sin (64 × 1)

=26.66V

(c)

To calculate the maximum rate of change of the magnetic flux through the rotating coil; we need to reflect on the equation for the maximum induced emf in terms of magnetic flux.

i.e $emf_max$ = $N\frac{d∅}{dt}$

since $emf_max$ = 29.66 and N = 44; we have:

29.66 = $44\frac{d∅}{dt}$

$\frac{d∅}{dt}$ = $\frac{29.66}{44}$

= 0.674 Wb/s

5 0

3 years ago

One of the important waste products of photosynthesis is oxygen gas. Where does this oxygen gas come from?

chubhunter [2.5K]

Answer:

End product of photosynthesis.

Explanation:

Photosynthesis is a process that plants undergo in the manufacture of their food. This is done in the presence of sunlight which is trapped by their coloring pigment called chlorophyll and reactants such as Carbon dioxide and Water.

6CO2+6H2O= C6H12O6+ 6O2

The oxygen gas which is a waste product of photosynthesis is released into the atmosphere and used by animals in respiration.

5 0

3 years ago

Read 2 more answers

A car moves round a circular track of radius 0.3m of two revolution per/sec find its angular velocity.

Pie

Answer:

the angular velocity of the car is 12.568 rad/s.

Explanation:

Given;

radius of the circular track, r = 0.3 m

number of revolutions per second made by the car, ω = 2 rev/s

The angular velocity of the car in radian per second is calculated as;

From the given data, we convert the angular velocity in revolution per second to radian per second.

$\omega = 2 \ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = 4\pi \ rad/s = 12.568 \ rad/s$

Therefore, the angular velocity of the car is 12.568 rad/s.

4 0

3 years ago

Determine the length of the object shown <br> 1.2 cm <br> 1.3 cm <br> 1.25 cm <br> 1.250 cm

stiks02 [169]

The correct answer is 1.25 because it is 1/2 of 1 1/2 and that is 1.25.

6 0

3 years ago

Read 2 more answers

A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o

Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

$f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz$

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

$\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m$

The fundamental frequency of the wire is given by

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N$

The tension in the wire is 67.9064 N

7 0

3 years ago