1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Mariulka [41]
3 years ago
5

Consider the four quantum numbers of an electron in an atom, n, l, ml, and ms. The energy of an electron in an isolated atom dep

ends on:______a. l, ml, and msonly.b. n only.c. n and l only.d. n, l, and mlonly.e. all four quantum numbers.
Physics
1 answer:
Effectus [21]3 years ago
6 0

Answer:

The energy of an electron in an isolated atom depends on b. n only.

Explanation:

The quantum number n, known as the principal quantum number represents the relative overall energy of each orbital.

The sets of orbitals with the same n value are often referred to as an electron shell, in an isolated atom all electrons in a subshell have exactly the same level of energy.

The principal quantum number comes from the solution of the Schrödinger wave equation, which describes energy in eigenstates E_n, and for the case of an hydrogen atom we have:

E_n=-\cfrac{13.6}{n^2}\, eV

Thus for each value of n we can describe the orbital and the energy corresponding to each electron on such orbital.

You might be interested in
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
atroni [7]

(a) 2.79 rev/s^2

The angular acceleration can be calculated by using the following equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha is the angular acceleration

\theta=50.0 rev is the number of revolutions made by the disk while accelerating

Solving the equation for \alpha, we find

\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

\theta=? is the number of revolutions made by the disk while accelerating

Solving the equation for \theta, we find

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

4 0
3 years ago
A flywheel with radius of 0.400 mm starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2rad/s2.
charle [14.2K]

Answer: 0.00024\ m/s^2

Explanation:

Given

Radius of flywheel is r=0.4\ mm

Angular acceleration \alpha=0.6\ rad/s^2

For no change in radius, tangential acceleration is  given as

\Rightarrow a_t=a\lpha \times r

Insert the values

\Rightarrow a_t=0.6\times 0.4\times 10^{-3}\ m/s^2\\\Rightarrow a_t=2.4\times 10^{-4}\ m/s^2\ \text{or}\ 0.00024\ m/s^2

5 0
3 years ago
The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give
Flauer [41]

Answer:

velocity = 1527.52 ft/s

Acceleration = 80.13 ft/s²

Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700(0.019)²)/sin66

a = (85 - 11.8047)/0.9135

a = 80.13 ft/s²

4 0
3 years ago
Aliens? do you believe?<br><br>To get the points go into detail :)
zzz [600]

Answer:

Yes

Explanation:

There are so many planets out there that there must be habitable planets if not in our galaxy but the Universe.

Although the chances of advanced life are slim, small primitive life like microbes or sea life may still exist.

7 0
3 years ago
Read 2 more answers
Which of the following types of weather data is collected in inches or centimeters?
aksik [14]
The correct answer is C) Rainfall
7 0
3 years ago
Read 2 more answers
Other questions:
  • If you push a cork into the neck of a bottle filled with air, what happens to the pressure inside the bottle?
    12·1 answer
  • Why is the study of black holes important?
    15·1 answer
  • A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns throu
    12·1 answer
  • How does the range change as the angle increases?
    7·1 answer
  • Do geochemists need to have a knowledge of physical science? Explain your<br> answer. please help:))
    7·1 answer
  • Objects fall at constant velocity
    12·2 answers
  • In the design of wall and column forms, the two most important factors are which of the following? a. rate of placement of the c
    7·1 answer
  • Lead has a density of 11. 5 g/cm³. A rectangular block of lead measures 7 cm x 5 cm x 2 cm.
    8·1 answer
  • What energy output objects work with the turbine
    14·1 answer
  • Select the correct answer.
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!