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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In order to answer this exercise you need to use the formulas
S = Vo*t + (1/2)*a*t^2
Vf = Vo + at
The data will be given as
Vf = final velocity = ?
Vo = initial velocity = 1.4 m/s
a = acceleration = 0.20 m/s^2
s = displacement = 100m
And now you do the following:
100 = 1.4t + (1/2)*0.2*t^2
t = 25.388s
and
Vf = 1.4 + 0.2(25.388)
Vf = 6.5 m/s
So the answer you are looking for is 6.5 m/s
Ultraviolet light with a wavelength shorter than visible light and a higher radiant energy than visible light.
The shorter the wavelength, the higher the frequency and energy.
The ratio of the maximum photoelectron kinetic energy to the work function will be 3:1.
<h3 /><h3>What is the photoelectric effect?</h3>
When a medium receives electromagnetic radiation, electrostatically charged particles are emitted from or inside it.
The emission of ions from a steel plate when light falls on it is a common definition of the effect. The substance could be a solid, liquid, or gas; and the released particles could be protons or electrons.
A particular metal emits photoelectrons when exposed to light with energy three times its work function:
The ratio of the maximum photoelectron kinetic energy to the work function will be;
Hence, the ratio of the maximum photoelectron kinetic energy to the work function will be 3:1.
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