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2 years ago

Which transition by an electron will release the greatest amount of energy? hurry!

Physics

2 answers:

Aleonysh [2.5K]2 years ago

8 0

Answer:

A is the answer

Explanation:

Sphinxa [80]2 years ago

5 0

highest energy level to the ground state.

Explanation:

The transition from the highest energy level to the ground state.

An electron has a discrete amount of energy accrued to it in any energy level it belongs to.

Electrons can transition between one energy level or the other.

When electrons change state, they either release or absorb energy.
When an atom absorbs energy, they move from their ground to final state which is consistent with the energy of the final state.
When electrons release energy, they move from excited state to their ground state.
Electrons will release the greatest amount of energy when they move from the highest energy level to the ground state.

Learn more:

Neil Bohr brainly.com/question/4986277

#learnwithBrainly

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3 years ago

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2 years ago

VP 3.12.1 Part APart complete A cyclist going around a circular track at 10.0 m/s has a centripetal acceleration of 5.00 m/s2. W

viktelen [127]

Answer:

A) r = 20.0 m

B) T = 41.6 s

C) = 6.1 m/s²

Explanation:

The centripetal acceleration is the one that explains that even though the cyclist is moving at a constant speed, his velocity is changing the direction all the time, keeping him around a circle.
This acceleration can be expressed as follows:

$a_{c} =\frac{v^{2}}{r} = \frac{(10.0m/s)^{2}}{r} = 5.00 m/s2 (1)$

Solving for r:

$r = \frac{v^{2}}{a_{c} } = \frac{(10.0m/s)^{2}}{5.00m/s2} = 20.0 m (2)$

We can apply the definition of linear velocity, remembering that the period is the time needed to complete an entire circle (T).
The arc around a circumference (the distance traveled) , is just 2*π*r, so applying the definition of linear velocity, we can write the following expression:

$v = \frac{\Delta s}{\Delta t} = \frac{2*\pi*r}{T} (3)$

Solving for T:

$T = \frac{\Delta s}{v} = \frac{2*\pi*r}{v} = \frac{2*\pi*265m}{40.0m/s} =41.6 s (4)$

The centripetal acceleration of the car from B) can be found as follows:

$a_{c} =\frac{v^{2}}{r} = \frac{(40.0m/s)^{2}}{265m} = 6.1 m/s2 (5)$

4 0

2 years ago