In the equation because that the kinetic power of an electron, the m represents mass, however how carry out we identify the mass?Example 1.5 of chapter 1 switch the speed of one electron emitted indigenous the surface ar of a sample of potassium (668 km/s) to Joules making use of Ek = 1/2 mv^2 by act the following:Ek = 1/2 x 9.109 x 10^-31 kg x 6.68 x 10^5 m/sso that the power in Joules is 2.03 x 10^-19 JMy inquiry is, whereby does the 9.109 x 10^-31 kg come from? go they use the AMU that potassium?


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Hey, the "m" friend are trying to find is in fact, "m(electron)" or the mass of the electron(s) gift shot at by a present of photons. The kinetic power equation shows the reality that if the power of the photon is higher than the energy required come displace the electron from its orbital, there will certainly be excess power released by the device for the electron to "pop" out, whereas if the power of the photon hitting the electron is less than the energy required to displace the electron from its orbital, the energy from the photon will just be dispersed.The adhering to is completely theoretical and unnecessary for the course, yet a inquiry I had actually prior to expertise the method photons functioned was that photons with power less 보다 the threshold energy are prefer streams the ping pong ball.s hitting a bowling sphere one through one. No issue how numerous photons fight the bowling ball, the sphere won"t move. Now a photon with energy greater than the threshold energy required, would certainly be a lot prefer a bowling round hitting a tennis ball, or even a tennis sphere of the exact same mass moving at high velocity come hit an additional tennis ball. Likewise remember the the position and momentum of one electron is almost indeterminate, so that bowling ball you"re aiming for through your ping pong ball can not in reality hit the bowling sphere you to be aiming for, but rather one more bowling round to the left. (kind of choose in bowling, where you"re aiming for all the pins, but you just strike one pin) sorry if this explanation go a little off track, but hopefully, it helped!
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Mohak Kumar 1LPosts: 11Joined: Fri Sep 26, 2014 2:02 afternoon

Re: Ek = 1/2mv^2....how execute we obtain the "m"?

Postby Mohak Kumar 1L » Thu Oct 09, 2014 7:13 pm


So the fixed of an electron is the very same regardless that what facet it is? 9.109 x 10^-31 kg is something that will certainly be given or execute we have to memorize it? ns asking because the evaluate of the photoelectric impact module had actually a trouble that forced the fixed of a sodium electron for the same equation. Also I to be wondering just how you acquired 10^-19 J from 10^-31 kg x 10^5 m/s, wouldn"t that be 10^-26?

Re: Ek = 1/2mv^2....how carry out we acquire the "m"?

Postby Niharika Reddy 1D » Thu Oct 09, 2014 10:30 afternoon


Mohak Kumar 1L wrote:So the mass of an electron is the very same regardless of what element it is? 9.109 x 10^-31 kg is something that will certainly be offered or perform we have to memorize it? im asking because the evaluate of the photoelectric impact module had actually a trouble that required the fixed of a sodium electron for the exact same equation. Additionally I to be wondering just how you obtained 10^-19 J native 10^-31 kg x 10^5 m/s, wouldn"t that be 10^-26?


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Yes, the massive of an electron is always the same. This value is offered to united state on the back of the routine table under the "Constants and also Equations" side (top right). Regarding your last question, the velocity should"ve to be squared due to the fact that Ek=1/2mv^2 (I think the was just a typo).