Answer:
the only element above is potassium
Answer:
The atomic structure of an atom involves 3 subatomic particles: the proton, neutron, and electron. The proton has a positive charge and is found in the core of the atom, with the neutral neutrons that also have a mass of 1 amu (atomic mass unit) just like the proton. The nucleus is the core of the atom and contains protons and neutrons and is practically the only area with mass. The electron cloud is basically an area surrounding the nucleus and it contains negative charged electrons. Electrons have no mass but are charged with a negative charge that keeps them. I really hope this helps :)
Explanation:
There is a helpful video that actually explains the structure of an atom in a rather fun way in just 2 minutes. It really does help big time and it's kinda funny if you look it up on YT and watch:
WKRP: Venus Explains the Atom
Have a wonderful great day :)
Apsidal precession—The major axis of Moon's elliptical orbit rotates by one complete revolution once every 8.85 years in the same direction as the Moon's rotation itself.
Answer : The energy released by an electron in a mercury atom to produce a photon of this light must be, ![4.56\times 10^{-19}J](https://tex.z-dn.net/?f=4.56%5Ctimes%2010%5E%7B-19%7DJ)
Explanation : Given,
Wavelength = ![435.8nm=435.8\times 10^{-9}m](https://tex.z-dn.net/?f=435.8nm%3D435.8%5Ctimes%2010%5E%7B-9%7Dm)
conversion used : ![1nm=10^{-9}m](https://tex.z-dn.net/?f=1nm%3D10%5E%7B-9%7Dm)
Formula used :
![E=h\times \nu](https://tex.z-dn.net/?f=E%3Dh%5Ctimes%20%5Cnu)
As, ![\nu=\frac{c}{\lambda}](https://tex.z-dn.net/?f=%5Cnu%3D%5Cfrac%7Bc%7D%7B%5Clambda%7D)
So, ![E=h\times \frac{c}{\lambda}](https://tex.z-dn.net/?f=E%3Dh%5Ctimes%20%5Cfrac%7Bc%7D%7B%5Clambda%7D)
where,
= frequency
h = Planck's constant = ![6.626\times 10^{-34}Js](https://tex.z-dn.net/?f=6.626%5Ctimes%2010%5E%7B-34%7DJs)
= wavelength = ![435.8\times 10^{-9}m](https://tex.z-dn.net/?f=435.8%5Ctimes%2010%5E%7B-9%7Dm)
c = speed of light = ![3\times 10^8m/s](https://tex.z-dn.net/?f=3%5Ctimes%2010%5E8m%2Fs)
Now put all the given values in the above formula, we get:
![E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}](https://tex.z-dn.net/?f=E%3D%286.626%5Ctimes%2010%5E%7B-34%7DJs%29%5Ctimes%20%5Cfrac%7B%283%5Ctimes%2010%5E%7B8%7Dm%2Fs%29%7D%7B%28435.8%5Ctimes%2010%5E%7B-9%7Dm%29%7D)
![E=4.56\times 10^{-19}J](https://tex.z-dn.net/?f=E%3D4.56%5Ctimes%2010%5E%7B-19%7DJ)
Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, ![4.56\times 10^{-19}J](https://tex.z-dn.net/?f=4.56%5Ctimes%2010%5E%7B-19%7DJ)
Answer:
4)experiments with cathode ray tubes
Explanation:
when sufficiently high voltage is applied across the electrods, current starts flowing through a stream of particles moving in the tube the negative electrode (cathode) to the positive electrode (anode). These were called Cathode Rays or Cathode Ray Particles