<span>Anger is to angry as fire is to blazing. </span>
Answer:
388.97 nm
Explanation:
The computation of the wavelength of this light in benzene is shown below:
As we know that
n (water) = 1.333
n (benzene) = 1.501
And, the wavelength of water is 438 nm
![\lambda (benzene) = \lambda (water) [\frac{n(water)}{n(benzene}]](https://tex.z-dn.net/?f=%5Clambda%20%28benzene%29%20%3D%20%5Clambda%20%28water%29%20%5B%5Cfrac%7Bn%28water%29%7D%7Bn%28benzene%7D%5D)
Now placing these values to the above formula
So,
= 388.97 nm
We simply applied the above formula so that we can easily determine the wavelength of this light in benzene could come
You should trust the primary source more.
This is because the primary source is make its conclusion from direct observation, while the secondary source is possibly making reference to another secondary source or to another primary.
The primary source should be trusted more because it is from direct observation.
We use the Rydberg Equation for this which is expressed as:
<span>1/ lambda = R [ 1/(n2)^2 - 1/(n1)^2]
</span>
where lambda is the wavelength, where n represents the final and initial states. Brackett series means that the initial orbit that electron was there is 4 and R is equal to 1.0979x10^7m<span>. Thus,
</span>
1/ lambda = R [ 1/(n2)^2 - 1/(n1)^2]
1/1.0979x10^7m = 1.0979x10^7m [ 1/(n2)^2 - 1/(4)^2]
Solving for n2, we obtain n=1.
All of the following
involve waves of electromagnetic energy except the rumble of thunder during a storm.
Electromagnetic waves<span> <span>are
used to transmit long/short/FM wavelength radio </span>waves, and TV/telephone/wireless signals or energies. They are
also responsible for transmiting energy in the form of microwaves, infrared radiation<span> (IR), visible light (VIS),
ultraviolet light (UV), X-rays, and gamma rays.</span></span>
The correct answer between all
the choices given is the second choice or letter B. I am hoping that this
answer has satisfied your query and it will be able to help you in your
endeavor, and if you would like, feel free to ask another question.
