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AysviL [449]
3 years ago
6

Waves must exhibit a constant phase difference for interference to occur. Please select the best answer from the choices provide

d T F
Physics
1 answer:
zalisa [80]3 years ago
8 0
True is what it is/ ,,,,,,,,,,,,,
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In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this
soldier1979 [14.2K]

Answer:

1). v = - 2960526m/s

2). Toward us

3). v = - 493421m/s

4). Toward us

5). v = 1480263m/s

6).  Away from us

7). v = 3207236m/s

8). Away from us

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest is 121.6 nm (\lambda_{0} = 121.6nm)

Redshift: \lambda_{measured} > \lambda_{0}

Blueshift: \lambda_{measured} < \lambda_{0}

Then, for this particular case it is gotten:

Star 1: \lambda_{measured} = 120.4nm

Star 2: \lambda_{measured} = 121.4nm

Star 3: \lambda_{measured} = 122.2nm

Star 4: \lambda_{measured} = 122.9nm

Star 1:

Blueshift: 120.4nm < 121.6nm

Toward us

Star 2:

Blueshift: 121.4nm < 121.6nm

Toward us

Star 3:

Redshift: 122.2nm > 121.6nm

Away from us

Star 4:

Redshift: 122.9nm > 121.6nm

Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

v = c\frac{\Delta \lambda}{\lambda_{0}}  (1)

Where \Delta \lambda is the wavelength shift, \lambda_{0} is the wavelength at rest, v is the velocity of the source and c is the speed of light.

v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}}) (2)

<em>Case for star 1 \lambda_{measured} = 120.4 nm:</em>

<em></em>

v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})

v = - 2960526m/s

Notice that the negative velocity means that is approaching to the observer.

<em>Case for star 2 \lambda_{measured} = 121.4 nm:</em>

v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})

v = - 493421m/s

<em>Case for star 3 \lambda_{measured} = 122.2 nm:</em>

v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})

v = 1480263m/s

<em>Case for star 4 \lambda_{measured} = 122.9 nm:</em>

v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})

v = 3207236m/s

4 0
3 years ago
Which of the following best describes the way that scientists make observations
horrorfan [7]
I don’t see any answer choice but the best way is asking questions about the natural phenomenon, making hypothesis, and predicting the consequences in the hypothesis.
6 0
3 years ago
When Cesium metal is illuminated with light of wavelength 300 nm, the photoelectrons emitted have a maximum kinetic energy of 2.
Allisa [31]

Answer:

i have no clue im sorry

Explanation:

8 0
3 years ago
How many total molecules are present in the formula below<br> 2H2O
Free_Kalibri [48]
There are 4 in there I believe
3 0
3 years ago
Read 2 more answers
A slit of width 0.01 mm has light of wavelength 600 nm passing through it onto a screen 60 cm away. how wide (in cm) is the cent
liubo4ka [24]

A slit of width 0.01 mm has light of wavelength 600 nm passing through it onto a screen 60 cm away.  The central maximum will be 7.2 cm .

The central maximum is the brightest central portion of the diffraction pattern. The central maximum is the widest and has the maximum intensity.

The diffraction pattern involves a central bright fringe, also known as the central maxima. Furthermore, central maxima is surrounded by dark and bright lines known as the secondary minima and maxima.

Given

slit width = 0.01 mm = .01 * 10^{-3} m

wavelength = 600 nm = 600 * 10^{-9} m

distance from screen = 60 cm = 60 * 10^{-2} m

width of central maxima = (2 * lambda * D) / d

                                        = 2 * 600 * 10^{-9} * 60 * 10^{-2}  / .01 * 10^{-3}

                                        = 7.2 * 10^{2} m

To learn more about central maxima here

brainly.com/question/14279157

#SPJ4

3 0
1 year ago
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