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choli [55]
3 years ago
13

The balmer series is formed by electron transitions in hydrogen that

Physics
1 answer:
RSB [31]3 years ago
4 0

Answer:

The Balmer series refers to the spectral lines of hydrogen, associated to the emission of photons when an electron in the hydrogen atom jumps from a level n \geq 3 to the level n=2.

The wavelength associated to each spectral line of the Balmer series is given by:

\frac{1}{\lambda}=R_H (\frac{1}{2^2}-\frac{1}{n^2})

where R_H is the Rydberg constant for hydrogen, and where n is the initial level of the electron that jumps to the level n = 2.

The first few spectral lines associated to this series are withing the visible part of the electromagnetic spectrum, and their wavelengths are:

656 nm (red, corresponding to the transition 3 \rightarrow 2)

486 nm (green, 4 \rightarrow 2)

434 nm (blue, 5 \rightarrow 2)

410 nm (violet, 6 \rightarrow 2)

All the following lines lie in the ultraviolet part of the spectrum. The limit of the Balmer series, corresponding to the transition \infty \rightarrow 2, is at 364.6 nm.

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Explain in your own words how the Doppler Effect is also applicable in our study of light.
arlik [135]

well in my own words, i'd saw the the doppler effect is similar to light because sound has a speed, and light does too.

so my theory is if you go fast enough everything would just become black, or maybe white? idk its hard to explain

but what my point is, is taht the doppler effect works in the same way, like if a car is moving towards you the sound is being emitted from the car and being pushed by the speed of the car making it have a much higher pitch, when the car is going away however it drops to a lower pitch due the the sound waves being DRAGGED by the car.

there hoped this helped I guess

8 0
3 years ago
A proton with charge 1.601019 As is moving at 2.4105 m/s through a magnetic field of 4.5 T. You want to find the force on the pr
Whitepunk [10]

The force on the proton is 17.4 N.

<h3>What is the force on the proton?</h3>

Now we know that the proton is positively charged and that the force on the charge as it moved through the magnetic field could be given by the relation; F = qvB

Where;

F = force

q = charge

v = velocity

B = magnetic field

Having said this, we can see that;

q = 1.601019 As or C

v =  2.4105 m/s

T = 4.5 T

F =  1.601019 As * 2.4105 m/s * 4.5 T

F = 17.4 N

Learn more about magnetic force:brainly.com/question/12824331

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6 0
2 years ago
1. A student lifts a box of books that weighs 185 N. The box is
aksik [14]

1)  148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

W = \Delta U = (mg) \Delta h

where

mg is the weight of the object

\Delta h is the change in height

For the box in this problem,

mg = 185 N

\Delta h = 0.800 m

Substituting into the equation, we find:

W=(185)(0.800)=148 J

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

W=Fd

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

W=(825)(35)=28875 J

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

W=Fd

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

W' = 2(28875)=57750 J

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

F = mg

where

m = 0.180 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

Solving the equation,

F=(0.180)(9.8)=1.76 N

Now we find the work done by gravity using the same formula applied before:

W=Fd

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

W=(1.76)(2.5)=4.4 J

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

\Delta h = 1.2 m

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

W=mg \Delta h

Solving for m, we find

m=\frac{W}{g \Delta h}

And substituting the numerical values, we find the mass of the box:

m=\frac{7000}{(9.8)(1.2)}=595.2 kg

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

W=mg \Delta h

Where \Delta h is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of \Delta h is the same for both of them, so the work they have done is exactly the same.

5 0
3 years ago
Categorize the particles according to whether they are found in the nucleus or the electron cloud.
inna [77]
1. neutral particles (neutrons) are in the nucleus
2. nucleus is in the nucleus
3. electron cloud is in the electron cloud
4. positively charged particles (protons) are in the nucleus
5. negatively charged particles (electrons) are in the electron cloud
4 0
3 years ago
Read 2 more answers
A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy o
Brrunno [24]

The electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.

<h3>Electric potential energy</h3>

When work is done on a positive test charge to move it from one location to another, potential energy increases and electric potential increases.

The electric potential energy between the charges when the second charge is at point b is calculated as follows;

ΔU = -w

Ui - Uf = w

Uf = Ui - w

where;

Uf is the final potential energy

Ui is the initial potential energy

w is the work done by the force

Uf = 5.4 x 10⁻⁸ J - (-1.9 x 10⁻⁸J)

Uf = 5.4 x 10⁻⁸ J + 1.9 x 10⁻⁸ J

Uf = 7.3 x 10⁻⁸ J

Thus, the electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.

Learn more about electric potential energy here: brainly.com/question/14306881

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7 0
2 years ago
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