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wariber [46]
3 years ago
11

The hydrogen atom, changing from its first excited state to its lowest energy state, emits light with a wavelength of 122 nm. Th

at is in the far ultraviolet. The sodium atom, which like hydrogen has one electron that gets excited outside a core of 10 other electrons, emits light at 589 nm making a similar transition from its first excited state to its lowest state. Which of these statements would be true about the sodium and hydrogen atoms and their spectra?
a. There would be other series of spectral lines ending on the first excited states of both atoms.
b. There would be a series of spectral lines in hydrogen with the longest wavelength one at 122 nm.
c. There would be a series of spectral lines in sodium with the longest wavelength one at 589 nm.
d. The hydrogen atom binds its electron more tightly than the sodium atom does, and would require more energy to remove its electron completely.
Physics
1 answer:
tatyana61 [14]3 years ago
8 0

Answer:

true b and c

Explanation:

n the electromechanical transitions of the atoms the relationship must be fulfilled

        \frac{x}{\lambda } = R (1 / nf - 1 / no²)

where for the final state nf = 1 giving in the case of hydrogen the Lymma series whose smallest wavelength is lam = 122 nm with nf = 1 and there are a series of spectral lines for each value of n of the final state

in the case of sodium so well it has a transition from an excited state to the kiss state (bad)

Now let's review the different proposals

a) False. The electronic potential for sodium is much lower than for hydrognosia

b) True

c) True

d) true

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DanielleElmas [232]

Answer: 363 Ω.

Explanation:

In a series AC circuit excited by a sinusoidal voltage source, the magnitude of the impedance is found to be as follows:

Z = √((R^2 )+〖(XL-XC)〗^2) (1)

In order to find the values for the inductive and capacitive reactances, as they depend on the frequency, we need first to find the voltage source frequency.

We are told that it has been set to 5.6 times the resonance frequency.

At resonance, the inductive and capacitive reactances are equal each other in magnitude, so from this relationship, we can find out the resonance frequency fo as follows:

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f = 1,600 Hz

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3 years ago
If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is
viktelen [127]
To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: V_{f}=V_{i}+gt
where
V_{f} is the final velocity 
V_{i} is the initial velocity 
g is the acceleration due to gravity 32 \frac{ft}{s^2}
t is the time 
2. Kinematic equation for distance: d=V_{i}t+ \frac{1}{2} gt^2
where
d is the distance 
V_{i} is the initial velocity 
V_{f} is the final velocity
g is the acceleration due to gravity 32 \frac{ft}{s^2}
t is the time 

First, we are going to convert 21.82 mi/h to ft/s:
21.82 \frac{mi}{h} =31.21 \frac{ft}{s}

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: V_{f}=0. We also know that it initial speed is 31.21 ft/s, so V_{i}=31.21. Lets replace those values in our formula to find t:
V_{f}=V_{i}+gt
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t= \frac{-31.21}{-32}
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Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
d=V_{i}t+ \frac{1}{2} gt^2
d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2
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Now we can add the height of the building and the maximum height of the object:
d=160+15.22=175.22ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
d=V_{i}t+ \frac{1}{2} gt^2
175.22=31.21t+ \frac{1}{2} (32)t^2
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Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
V_{f}=V_{i}+gt
V_{f}=31.21+(32)(2.47)
V_{f}=110.25 ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s


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