The wavelength of the third line in the Lyman series, and identify the type of EM radiation
In this series, the spectral lines are obtained when an electron makes a transition from any high energy level (n=2,3,4,5... ). The wavelength of light emitted in this series lies in the ultraviolet region of the electromagnetic spectrum.
1 / lambda = R(h)* ( 
- 
)
= 109678 ( 
- 
)
= 109678 (8/9)
Lambda = 9 / (109678 * 8 )
= 102.6 * 
m = 102.6 nm
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To answer the following questions for this specific problem:
a. 11.48 secs
b. Vp = a*t*3.6 =
3*11.48*3.6 = 124.0 km/h
<span>c. 9.1 secs. </span>
I am hoping that this answer has satisfied your query about
and it will be able to help you.
Answer:
The energy level is 5.
Explanation:
Given that,
n = 3
l = 2
We know that,
l shows the number of sub-shells and define the number of angular nodes.
n shows the number of electron shell.
is a quantum number. It is define the number of energy level in a sub-shells .
is define the spin of the electron.
So, The quantum number is
is 
and 
for every energy level.
The energy level is 5.
Hence, The energy level is 5.
Answer:
the pressure at B is 527psf
Explanation:
Angular velocity, ω = v / r
ω = 20 /1.5
= 13.333 rad/s
Flow equation from point A to B
![P_A+rz_A-\frac{1}{2} Pr_A^2w^2=P_B+rz_B-\frac{1}{2} pr^2_Bw^2\\\\P_B = P_A + r(z_A-z_B)+\frac{1}{2} pw^2[(r_B^2)-(r_A)^2]\\\\P_B = [25 +(0.8+62.4)(0-1)+\frac{1}{2}(0.8\times1.94)\times(13.333)^2[2.5^2-1.5^2] ]\\\\P_B = 25 - 49.92+551.79\\\\P_B = 526.87psf\\\approx527psf](https://tex.z-dn.net/?f=P_A%2Brz_A-%5Cfrac%7B1%7D%7B2%7D%20Pr_A%5E2w%5E2%3DP_B%2Brz_B-%5Cfrac%7B1%7D%7B2%7D%20pr%5E2_Bw%5E2%5C%5C%5C%5CP_B%20%3D%20P_A%20%2B%20r%28z_A-z_B%29%2B%5Cfrac%7B1%7D%7B2%7D%20pw%5E2%5B%28r_B%5E2%29-%28r_A%29%5E2%5D%5C%5C%5C%5CP_B%20%3D%20%5B25%20%2B%280.8%2B62.4%29%280-1%29%2B%5Cfrac%7B1%7D%7B2%7D%280.8%5Ctimes1.94%29%5Ctimes%2813.333%29%5E2%5B2.5%5E2-1.5%5E2%5D%20%20%5D%5C%5C%5C%5CP_B%20%3D%2025%20-%2049.92%2B551.79%5C%5C%5C%5CP_B%20%3D%20526.87psf%5C%5C%5Capprox527psf)
the pressure at B is 527psf
A piece of blue paper appears blue because the paper absorbs all colors of light except blue. <em> (b)</em>
So any light that bounces off of the paper and enters your eye must be blue light !
