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dedylja [7]
3 years ago
9

True or false in order for light to be emitted electrons must move from high energy state to a lower energy state?

Physics
2 answers:
lara31 [8.8K]3 years ago
5 0

Answer:

B

I think it is false.

It does not move to lower energy state.

victus00 [196]3 years ago
5 0
The answer to your question is true
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In an open system such as a campfire matter can
eimsori [14]
In an open system such as a campfire, matter can lose particles, gain particles or exchange particles.
4 0
3 years ago
A baseball hit just above the ground leaves the bat 27 m/s at 45° above the horizontal. A) How far away does the ball strike the
Sedbober [7]

Answer:

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

           v = u + at

           0 = 19.09 - 9.81 t

            t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

     Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

     So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

              S = ut + 0.5 at²

              H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

    Maximum height of the ball = 18.57 m

6 0
3 years ago
(a) What is the ionization energy of a hydrogen atom that is in the n = 6 excited state? (b) For a hydrogen atom, determine the
crimeas [40]

Answer:

(a) 0.3778 eV

(b) Ratio = 0.0278

Explanation:

The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

E=\frac {-13.6}{n^2}\ eV

(a) The energy of the electron in n= 6 excited state is:

E=\frac {-13.6}{6^2}\ eV

E=-0.3778\ eV

Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV

(b) For first orbit energy is:

E=\frac {-13.6}{1^2}\ eV

E=-13.6\ eV

Ratio=\frac {E_6}{E_1}

Ratio=\frac {-0.3778}{-13.6}

Ratio = 0.0278

7 0
3 years ago
An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accele
kirza4 [7]

Answer:

W = 462.5 keV

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

F = qE

F = (1.6 \times 10^{-19})(1.85 \times 10^6)

F = 2.96 \times 10^{-13} N

now the distance moved by the electron is given as

d = 0.25 m

so we have

W = F.d

W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)

W = 7.4 \times 10^{-14} J

now we have to convert it into keV units

so we have

1 keV = 1.6 \times 10^{-16} J

W = 462.5 keV

5 0
3 years ago
What is a free electron? Can someone help?
Aleksandr [31]
A free electron is one which has become detached from a covalent bond between two atoms and is able to move around from atom to atom and possibly take part in electric current flow.
3 0
3 years ago
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