Answer:
486nm
Explanation:
in order for an electron to transit from one level to another, the wavelength emitted is given by Rydberg Equation which states that
![\frac{1}{wavelength}=R.[\frac{1}{n_{f}^{2} } -\frac{1}{n_{i}^{2} }] \\n_{f}=2\\n_{i}=4\\R=Rydberg constant =1.097*10^{7}m^{-1}\\subtitiute \\\frac{1}{wavelength}=1.097*10^{7}[\frac{1}{2^{2} } -\frac{1}{4^{2}}]\\\frac{1}{wavelength}= 1.097*10^{7}*0.1875\\\frac{1}{wavelength}= 2.06*10^{6}\\wavelength=4.86*10{-7}m\\wavelength= 486nm\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bwavelength%7D%3DR.%5B%5Cfrac%7B1%7D%7Bn_%7Bf%7D%5E%7B2%7D%20%7D%20-%5Cfrac%7B1%7D%7Bn_%7Bi%7D%5E%7B2%7D%20%7D%5D%20%5C%5Cn_%7Bf%7D%3D2%5C%5Cn_%7Bi%7D%3D4%5C%5CR%3DRydberg%20constant%20%3D1.097%2A10%5E%7B7%7Dm%5E%7B-1%7D%5C%5Csubtitiute%20%5C%5C%5Cfrac%7B1%7D%7Bwavelength%7D%3D1.097%2A10%5E%7B7%7D%5B%5Cfrac%7B1%7D%7B2%5E%7B2%7D%20%7D%20-%5Cfrac%7B1%7D%7B4%5E%7B2%7D%7D%5D%5C%5C%5Cfrac%7B1%7D%7Bwavelength%7D%3D%201.097%2A10%5E%7B7%7D%2A0.1875%5C%5C%5Cfrac%7B1%7D%7Bwavelength%7D%3D%202.06%2A10%5E%7B6%7D%5C%5Cwavelength%3D4.86%2A10%7B-7%7Dm%5C%5Cwavelength%3D%20486nm%5C%5C)
Hence the photon must possess a wavelength of 486nm in order to send the electron to the n=4 state
Using
F= mv²/r
4 = 0.5×v² / 2
8 /0.5 = v²
v²=16
v= √16
v= 4 ms-¹
Answer:
"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2."
Explanation:
The full question has not been provided, so I just copied this into the web and found this answer and explanation on quizlet:
"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2.
D sin θ = m λ
if the wavelengths are the same, then if the angle is smaller, the slit width must be larger. The top photo shows a pattern that is more closely spaced. That means the angle is smaller. The slit width must be larger."
This answer/explanation should be correct, as we are looking at bright fringes and the formula being used corresponds to the parameters of the question.
Hope this helps!
in china, there is a family limit for only having 1 child
at 10 billion people on earth, we will most likely run out of food supply
The trickiest part of this problem was making sure where the Yakima Valley is.
OK so it's generally around the city of the same name in Washington State.
Just for a place to work with, I picked the Yakima Valley Junior College, at the
corner of W Nob Hill Blvd and S16th Ave in Yakima. The latitude in the middle
of that intersection is 46.585° North. <u>That's</u> the number we need.
Here's how I would do it:
-- The altitude of the due-south point on the celestial equator is always
(90° - latitude), no matter what the date or time of day.
-- The highest above the celestial equator that the ecliptic ever gets
is about 23.5°.
-- The mean inclination of the moon's orbit to the ecliptic is 5.14°, so
that's the highest above the ecliptic that the moon can ever appear
in the sky.
This sets the limit of the highest in the sky that the moon can ever appear.
90° - 46.585° + 23.5° + 5.14° = 72.1° above the horizon .
That doesn't happen regularly. It would depend on everything coming
together at the same time ... the moon happens to be at the point in its
orbit that's 5.14° above ==> (the point on the ecliptic that's 23.5° above
the celestial equator).
Depending on the time of year, that can be any time of the day or night.
The most striking combination is at midnight, within a day or two of the
Winter solstice, when the moon happens to be full.
In general, the Full Moon closest to the Winter solstice is going to be
the moon highest in the sky. Then it's going to be somewhere near
67° above the horizon at midnight.
