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[Ended]Physics Tutoring Session #5 Energy Quantization on Wed. 24th Feb 2016 @ 3:00 - 4:00pm
Physics
Chima Sunday (Tutor)
24-02-2016 11:55:00 +0000Good day all
Today we will be tutoring on Energy Quantization. The concept of Energy Quantization is the most fundamental in the study of modern Physics, it is often tested in JAMB and other exams. so lets take a quick look into this topics.
In 1910, Rutherford told us that the J.J Thompson model of the atom was wrong by coming up with his heavy central positively nucleus with electron circling around it. But since he said that the nucleus is positive, electron negative, so coulombs force should push the electron inside the nucleus. But this do not happen. That was the doom of Rutherford until Niels Bohr stepped in 1913 to tell him that look o, this electrons you are talking about circles the nucleus in a discrete energy level called quantum level. Actually Neils Bohr borrowed this idea from Max Planck theory of 1900 of photons quantization.
This Energy level is given thus
En = nhf
n = quantum numbers. n = 1,2,3,........
h = planck constant = 6.6*10-34J.s
f = frquency of the orbiting electron.
recall that from waves that velocity = frequency times wavelength.
then f = c/ λ
so En = nhc/λ
c = speed of light
lahmda(λ) = wavelength.
**Note that this session begins from 3:00 - 4:00pm. Before the session begins, you can put down your questions on paper and then ask them when the session officially begins by 3:00pm on any area in physics while giving more consideration on the above topic.
This session has ended
Thanks to every one that participated especially @Femmy Adewale a big thanks to you.
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Chima Sunday (Tutor)
Ok another day is here. please do ask your questions, discuss your challenges, and show appreciation when your questions are answered to your satisfaction. welcome once again.
0 24-02-2016 14:03:00 +0000
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Chima Sunday (Tutor)
let us start with this one. can any one try? JAMB 2015 (Question 21)What is the energy with which an incident light of wave length 10^-7m falls on a metal and electrons are emitted? [c = 3.0 x 10^8ms-1; h = 6.63 x 10^-34Js] Option A 1.99 x 10^-18J Option B 1.99 x 10^-16J Option C 1.99 x 10^-19J Option D 1.99 x 10^-20J Answer Option A Solution Wavelength λ = 10^-7 Velocity of light c = 3 × 10^8ms-1 Planck constant = 6.63 × 10^-34Js Energy of incident light, E = hf Where f is the frequency E = hc/λ recall 〈c = λ × v〉 E = (6.63 × 10^-34 × 3 × 10^8)/ 10^-7 E = 19.89 × 10^(-34 + 8 +7) E = 1.989 × 10 × 10^-19 E = 1.989 × 10^-18J = 9 x 10^-16J
0 24-02-2016 14:05:00 +0000
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Chima Sunday (Tutor)
Also note you are not only restricted to ask question on this only. you can equally ask question on any area in physics.
0 24-02-2016 14:08:00 +0000
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Chima Sunday (Tutor)
Hello class! hope we are together. Well this another one to try JAMB 2013 (Question 47)
In photo-emission, the number of photo electrons ejected per second depends on the
Option A frequency of the beam
Option B work function of the metal
Option C intensity of the beam
Option D threshold frequency of the metal
Answer is Option C The number of photo electrons emitted per second is directly proportional to the intensity of radiating beam The energy of the emitted electrons depends on the frequency of the incident radiating beam. The threshold frequency is that frequency in which below it no electrons are emitted.0 24-02-2016 14:16:00 +0000
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