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2 years ago

Which states of a hydrogen atom can be excited by a collision with an electron with kinetic energy k = 12. 5 ev?

Physics

1 answer:

Sophie [7]2 years ago

5 0

The answer is the excited state of the hydrogen atom, n=24 and Z=23

Kinetic energy of electron is, $\mathrm{KE}=\frac{13.6 \mathrm{Z}^{2}}{\mathrm{n}^{2}} \mathrm{eV}$

For the excited state of the hydrogen atom , n=24 and Z=23

K.E= $\frac{13.6 \times 23^{2}}{24^{2}}=$ 12.5

What is the hydrogen atom?

The hydrogen atom is the most basic of all the atoms since it just has one proton and one electron.
There are two more hydrogen isotopes, deuterium and tritium, in addition to protium, which is the most prevalent form of the atom.
The hydrogen atom consists of a proton and an electron, and has a spherical symmetry that can most easily be studied using a spherical polar coordinate frame.

To learn more about hydrogen atom visit: brainly.com/question/1462347?

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How much work is done by the force lifting a 0.1-kilogram hamburger vertically upward at a constant velocity 0.3 meter from a ta

pychu [463]

Answer:

0.2943 Nm

Explanation:

Work done is given a the product of force and diatance moved and expressed by the formula

W=Fd

Here W represent work, F is applied force and d is perpendicular distance

Also, we know that F=mg where m is the mass of an object and g is acceleration due to gravity. Substituting this back into the initial equation then

W=mgd

Taking acceleration due to gravity as 9.81 m/s2 and substituting mass with 0.1 kg and distance with 0.3 m then

W=0.1*9.81*0.3=0.2943 Nm

5 0

3 years ago

Then it is called

DiKsa [7]

Answer:

It is called force of friction

Explanation:

The force of friction is a force that acts between two objects whose surfaces are in contact with each other.

Consider the typical case of an object sliding along a certain surface. There are two types of frictions:

- Static friction: this is the force of friction that acts when the object is not in motion yet. If you push the object forward with a force F, the object will not move immediately, but it will "oppose" to this motion with a force of static friction exactly equal to the push applied:

$F_f = F$

However, this force of static friction has a maximum value, which is given by

$F_{max} = \mu_s N$

where

$\mu_s$ is the coefficient of static friction

N is the normal reaction exerted by the surface on the object

So, when $F$ becomes greater than $F_{max}$ , the static friction is no longer able to balance the push applied, and the object will start sliding forward.

- Kinetic friction: this is the force of friction that acts when the object is already in motion. Its magnitude is given by

$F_f = \mu_k N$

where

$\mu_k$ is the coefficient of kinetic friction, and its value is generally smaller than $\mu_s$ . The direction of this force is also opposite to the direction of motion of the object.

8 0

3 years ago

A 102 kg football player runs at a speed of 8 m/s to sack the quarterback. What is

leva [86]

The mass of the quarterback is 61.2 kg.

Explanation:

mass of the football player = m1 = 102 kg

mass of the quarterback = m2 = ?

velocity of the football player = v1 = 8 m/s

According to the law of conservation of momentum:

The total momentum of a system before and after the collision remains constant. Assuming the situation as an isolated system which is not affected by any external factors, we have:

m₁v₁ + m₂v₂ = (m₁+m₂)V

Here, we need to find m₂.

We assume that the quarterback is standing still when he is attacked by the football player so v₂ = 0 m/s

After the collision both of them fall to the ground with a velocity of 5 m/s so V = 5 m/s

$102(8) + m2(0) = (102 + m2)(5)\\816 + 0 = (102 + m2)(5)\\816/5 = 102 + m2\\163.2 - 102 = m2\\m2 = 61.2 kg$

Keywords: momentum, velocity, law of conservation of momentum

Learn more about Law of Conservation of Momentum from brainly.com/question/7538238

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3 0

3 years ago

A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,

n200080 [17]

Answer:

a. $t = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$ b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

$t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0} +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁ = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

6 0

3 years ago

What is the speed of sound in air at 40°C?

Blababa [14]

343 meters per second Is the answer

4 0

3 years ago

Read 2 more answers