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Serhud [2]
3 years ago
5

If a hydrogen atom has its electron in the n= 4 state, how much energy in eV is needed to ionize it?

Physics
1 answer:
ololo11 [35]3 years ago
4 0

Answer:

Explanation:

Energy of electron in n = 4 state

= - 13.6/4² =- 0.85 eV

So energy needed to extract it from atom = .85 eV.

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What is released when an excited electron returns to a lower energy state?
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A proton

Explanation:

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How does the planet Venus look to a person standing on the Earth?
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A person would look like an ant if a human was standing next to Venus

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3 years ago
In the model of the hydrogen atom due to Niels Bohr, the electron moves around the proton at a speed of 3.3 × 106 m/s in a circl
irga5000 [103]

Answer:

1.5048\times 10^{-23}\ Am^2

Explanation:

q = Charge of proton = 1.6\times 10^{-19}\ C

r = Radius of circle = 5.7\times 10^{-11}\ m

v = Velocity of proton = 3.3\times 10^6\ m/s

Magnetic moment is given by

M=\frac{1}{2}qrv\\\Rightarrow M=\frac{1}{2}1.6\times 10^{-19}\times 5.7\times 10^{-11}\times 3.3\times 10^6\\\Rightarrow M=1.5048\times 10^{-23}\ Am^2

The magnetic moment associated with this motion is 1.5048\times 10^{-23}\ Am^2

5 0
3 years ago
Based on the law of conversation of energy how can we reasonably improve a machines ability to do work?
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MARK ME BRAINLIEST!!

your answer should be “C”.
4 0
3 years ago
Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin
likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma    (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0     and T - mgcos57 - F = m(0) = 0

Mg - T = 0    (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F  

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0
3 years ago
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