The answer is n= 6.
What is Balmer series?
The Balmer series is the portion of the emission spectrum of hydrogen that represents electron transitions from energy levels n > 2 to n = 2. These are four lines in the visible spectrum. They are also known as the Balmer lines. The four visible Balmer lines of hydrogen appear at 410 nm, 434 nm, 486 nm and 656 nm.
For the Balmer series, the final energy level is always n=2. So, the wavelengths 653.6, 486.1, 434.0, and 410.2 nm correspond to n=3, n=4, n=5, and n=6 respectively. Since the last wavelength, 410.2 nm, corresponds to n=6, the next wavelength should logically correspond to n=7.
To solve for the wavelength, calculate the individual energies, E2 and E7, using E=-hR/(n^2). Then, calculate the energy difference between E2 (which is the final) and E7 (which is the initial). Finally, use lamba=hc/E to get the wavelength.
To learn more about emission spectrum click on the link below:
brainly.com/question/24213957
#SPJ4
Insulin and glucagon help maintain blood glucose levels
Answer:
42244138.951 m
Explanation:
G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²
r = Radius of orbit from center of earth
M = Mass of Earth = 5.98 × 10²⁴ kg
m = Mass of Satellite
The satellite revolves around the Earth at a constant speed
Speed = Distance / Time
The distance is the perimeter of the orbit
![v=\frac{2\pi \times r}{24\times 3600}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2%5Cpi%20%5Ctimes%20r%7D%7B24%5Ctimes%203600%7D)
The Centripetal force of the satellite is balanced by the universal gravitational force
![m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m](https://tex.z-dn.net/?f=m%5Cfrac%7Bv%5E2%7D%7Br%7D%3D%5Cfrac%7BGMm%7D%7Br%5E2%7D%5C%5C%5CRightarrow%20%5Cfrac%7B%5Cleft%28%5Cfrac%7B2%5Cpi%20%5Ctimes%20r%7D%7B24%5Ctimes%203600%7D%5Cright%29%5E2%7D%7Br%7D%3D%5Cfrac%7B6.667%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.98%5Ctimes%2010%5E%7B24%7D%7D%7Br%5E2%7D%5C%5C%5CRightarrow%20%5Cleft%28%5Cfrac%7B2%5Cpi%20%5Ctimes%20r%7D%7B24%5Ctimes%203600%7D%5Cright%29%5E2%3D6.667%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.98%5Ctimes%2010%5E%7B24%7D%5C%5C%5CRightarrow%20r%5E3%3D%5Cfrac%7B6.667%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.98%5Ctimes%2010%5E%7B24%7D%5Ctimes%20%2824%5Ctimes%203600%29%5E2%7D%7B%282%5Cpi%29%5E2%7D%5C%5C%5CRightarrow%20r%3D%5Cleft%28%5Cfrac%7B6.667%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.98%5Ctimes%2010%5E%7B24%7D%5Ctimes%20%2824%5Ctimes%203600%29%5E2%7D%7B%282%5Cpi%29%5E2%7D%5Cright%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5C%5C%5CRightarrow%20r%3D42244138.951%5C%20m)
The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m
a). for velocity, you must have a number, a unit, and a direction.
Yes. This one isn't bad. The 'number' and the 'unit' are the speed.
b). the si units for velocity are miles per hour.
No. That's silly.
'miles' is not an SI unit, and 'miles per hour'
is only a speed, not a velocity.
c). the symbol for velocity is .
You can use any symbol you want for velocity, as long as
you make its meaning very clear, so that everybody knows
what symbol you're using for velocity.
But this choice-c is still wrong, because either it's incomplete,
or else it's using 'space' for velocity, which is a very poor symbol.
d). to calculate velocity, divide the displacement by time.
Yes, that's OK, but you have to remember that the displacement
has a direction, and so does the velocity.