PS Final Exam

1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True B. False

Studentka2010 [4]

1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True

2. The maximum number of electrons allowed in a p sublevel of the 3rd principal level is?
B.6

3. A neutral atom has a ground state electronic configuration of 1s^2 2s^2. Which of the following statements concerning this atom is/are correct?
B. All of the above.

4 0

3 years ago

Read 2 more answers

Compute the velocity of an electron that has been accelerated through a difference of potential of 100 volts. express your answe

Elodia [21]

The velocity of an electron that has been accelerated through a difference of potential of 100 volts will be 5.93 * $10^{6}$ m/s

Electrons move because they get pushed by some external force. There are several energy sources that can force electrons to move. Voltage is the amount of push or pressure that is being applied to the electrons.

By conservation of energy, the kinetic energy has to equal the change in potential energy, so KE=q*V. The energy of the electron in electron-volts is numerically the same as the voltage between the plates.

given

charge of electron = 1.6 × $10^{-19}$ C

mass of electron = 9.1 × $10^{-31}$ kg

Force in an electric field = q*E

potential energy is stored in the form of work done

potential energy = work done = Force * displacement

= q * (E * d)

= q * (V) = 1.6 × $10^{-19}$ * 100

stored potential energy = kinetic energy in electric field

kinetic energy = 1/2 * m * $v^{2}$

= 1/2 * 9.1 × $10^{-31}$ * $v^{2}$

equation both the equations

1/2 * 9.1 × $10^{-31}$ * $v^{2}$ = 1.6 × $10^{-17}$

$v^{2}$ = 0.352 * $10^{14}$ m/s

$v^{2}$ = 35.2 * $10^{12}$

= 5.93 * $10^{6}$ m/s

To learn more about kinetic energy in electric field here

brainly.com/question/8666051

#SPJ4

3 0

11 months ago

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

4 0

3 years ago

Star #1 is approaching the Earth with speed v. Star #2 is receding from the Earth with the same speed v. Measurements of the sam

leva [86]

Answer:

b. Both stars will have the same shift.

Explanation:

It's a very simple problem to solve. Star 1 is approaching toward Earth with a speed v, so let's assume that the change in Doppler Shift is +F and Star 2 is moving away so the change in Doppler shift is -F. But it's time to notice the speed of both stars and that is same but only directions are different. speed is the main factor here. The magnitude of both shifts is F as we can see and + and - are showing there direction of motion. So, because of same amount of speed, both stars will have same shift magnitude. (Just the directions are different)

4 0

3 years ago

A very narrow beam of white light is incident at 40.80° onto the top surface of a rectangular block of flint glass 11.6 cm thick

DerKrebs [107]

Dispersion angle = 0.3875 degrees.
Width at bottom of block = 0.09297 cm
Thickness of rainbow = 0.07038 cm
Snell's law provides the formula that describes the refraction of light. It is:
n1*sin(Î¸1) = n2*sin(Î¸2)
where
n1, n2 = indexes of refraction for the different mediums
Î¸1, Î¸2 = angle of incident rays as measured from the normal to the surface.
Solving for Î¸2, we get
n1*sin(Î¸1) = n2*sin(Î¸2)
n1*sin(Î¸1)/n2 = sin(Î¸2)
asin(n1*sin(Î¸1)/n2) = Î¸2
The index of refraction for air is 1.00029, So let's first calculate the angles of the red and violet rays.
Red:
asin(n1*sin(Î¸1)/n2) = Î¸2
asin(1.00029*sin(40.80)/1.641) = Î¸2
asin(1.00029*0.653420604/1.641) = Î¸2
asin(0.398299876) = Î¸2
23.47193844 = Î¸2
Violet:
asin(n1*sin(Î¸1)/n2) = Î¸2
asin(1.00029*sin(40.80)/1.667) = Î¸2
asin(1.00029*0.653420604/1.667) = Î¸2
asin(0.39208764) = Î¸2
23.08446098 = Î¸2
So the dispersion angle is:
23.47193844 - 23.08446098 = 0.38747746 degrees.
Now to determine the width of the beam at the bottom of the glass block, we need to calculate the difference in the length of the opposite side of two right triangles. Both triangles will have a height of 11.6 cm and one of them will have an angle of 23.47193844 degrees, while the other will have an angle of 23.08446098 degrees. The idea trig function to use will be tangent, where
tan(Î¸) = X/11.6
11.6*tan(Î¸) = X
So for Red:
11.6*tan(Î¸) = X
11.6*tan(23.47193844) = X
11.6*0.434230136 = X
5.037069579 = X
And violet:
11.6*tan(Î¸) = X
11.6*tan(23.08446098) = X
11.6*0.426215635 = X
4.944101361 = X
So the width as measured from the bottom of the block is: 5.037069579 cm - 4.944101361 cm = 0.092968218 cm
The actual width of the beam after it exits the flint glass block will be thinner. The beam will exit at an angle of 40.80 degrees and we need to calculate the length of the sides of a 40.80/49.20/90 right triangle. If you draw the beams, you'll realize that:
cos(Î¸) = X/0.092968218
0.092968218*cos(Î¸) = X
0.092968218*cos(40.80) = X
0.092968218*0.756995056 = X
0.070376481 = X
So the distance between the red and violet rays is 0.07038 cm.

7 0

2 years ago