{"id":4097,"date":"2023-10-06T09:36:23","date_gmt":"2023-10-06T09:36:23","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=4097"},"modified":"2023-10-06T09:38:58","modified_gmt":"2023-10-06T09:38:58","slug":"week-7-ss3-second-term-physics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-7-ss3-second-term-physics-notes\/","title":{"rendered":"Week 7 &#8211; SS3 Second Term Physics Notes"},"content":{"rendered":"<p>\u00a0<strong>WEEK SEVEN<br \/>\n<\/strong><strong>TOPIC: PHOTO ELECTRIC EFFECT<br \/>\n<\/strong><strong>CONTENT<br \/>\n<\/strong><\/p>\n<ul>\n<li>Application<strong><br \/>\n\t\t\t<\/strong><\/li>\n<li>Photoelectric Equation\n<\/li>\n<li>Threshold frequency\n<\/li>\n<li>Work funtion\n<\/li>\n<li>x-ray\n<\/li>\n<\/ul>\n<p>When light falls on a metal surface, electrons are emitted, this process is called photo electric effect emission,  the emitted electrons are known as photo electrons.<br \/>\nThe maximum kinetic energy of the photo electrons are independent of the intensity of the incident light but depends on the frequency or wavelength of the incident light.<br \/>\n<img decoding=\"async\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0936_Week7SS3Se1.jpg\" alt=\"\"\/><br \/>\n\tIncreasing the intensity of light increases the number of photo electron but does not increase the energy or velocity. The absorbed energy is used to overcome the potential barrier of the  photo electrons.<\/p>\n<p>\u00a0<strong>APPLICATION<br \/>\n<\/strong>Photoelectric emissions is  used in the following :<br \/>\n     I Burglary alarm<br \/>\nii Television camera<br \/>\niii Automatic devices for switching light at dusk e. street light.<br \/>\niv. Sound production of film track<br \/>\nv. industrial controls and counting operations.<\/p>\n<p>\u00a0<strong>EINSTEN  PHOTOELECTRIC EQUATION<br \/>\n<\/strong>Einstein photoelectric equation is given  by<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0E = hf \u2013 w<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0W = hfo<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0E = maximum kinetic energy that can be given to a photo electrons<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0W = work function<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0fo = Threshold frequency<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0hf = W = maximum energy of the liberated. Photoelectrons.<\/p>\n<p>\u00a0<strong>THRESHOLD FREQUENCY (fo)<br \/>\n<\/strong>This is the lowest frequency that can cause photo emission of electrons from a metallic surface. Below threshold frequency, emission will not occur.<\/p>\n<p>\u00a0<strong>WORK FUNCTION  (W = hfo)<br \/>\n<\/strong>This is the minimum energy required to liberate electrons from a metallic surface.<br \/>\nW = hfo.<\/p>\n<p>\u00a0Example<br \/>\nCompute the frequency of the photon whose energy is required to eject a surface electron with a kinetic energy of 3.5 x 10-16 eV if the work function of the metal is 3.0 x 10-16 eV<br \/>\n\u00a0\u00a0\u00a0\u00a0(h = 6.6 x 10<sup>-34<\/sup>JS, 1eV = 1.6 x 10<sup>-19<\/sup>J ).<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0E = hf \u2013 w<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0E + w = hf<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0E + W  = f<br \/>\n               H<br \/>\n\u00a0\u00a0\u00a0\u00a0   =  ( 3.5 + 3.0) x 10<sup>-16 <\/sup>x 1.6 X 10<sup>-19<\/sup><\/p>\n<ol>\n<li>x 10<sup>-34<\/sup>\n\t\t<\/li>\n<\/ol>\n<p>\u00a0\u00a0\u00a0\u00a0  =   6.5 x 1.6 x 10<sup>-16 -19+ 34<\/sup><br \/>\n\t\t                   6.6<br \/>\n\u00a0\u00a0\u00a0\u00a0=  1.58 x 10<sup>-1<\/sup>  Hz<\/p>\n<p>\u00a0<strong>THRESHOLD WAVELENGTH<br \/>\n<\/strong>The threshold wavelength is the longest wavelength that will produce photo electrons when the<br \/>\nsurface is illuminated.<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0W = hfo<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0W = hc<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0       \u03bb<strong><sub>o<\/sub><\/strong><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u03bb<strong><sub>o<\/sub><\/strong> =hc<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0      W<br \/>\nThe work frequency of Lithium is 2.30eV, calculate <\/p>\n<ol>\n<li>the maximum energy in Joules of photoelectrons liberated by light of wavelength 3.3 x 10<sup>-17<\/sup>m\n<\/li>\n<li>the threshold wavelength of the metal.\n<\/li>\n<\/ol>\n<p>\u00a0<\/p>\n<ol>\n<li>W = 2.3 ev\n<\/li>\n<\/ol>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0E = hf \u2013 w<br \/>\n\u00a0\u00a0\u00a0\u00a0  \u00a0\u00a0\u00a0\u00a0   =  hc    &#8211; w<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0        \u03bb<\/p>\n<ol>\n<li>\n<div>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0   =   6.6 x 10<sup>-34<\/sup> x 3.0 x 10<sup>8<\/sup> \u2013 (2.3 x 1.6 x 10<sup>-19<\/sup>)\n<\/div>\n<p>   \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03.3 x 10<sup>-17<\/sup><br \/>\n\t\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 2.208 x 10<sup>-27<\/sup>J <\/p>\n<p>\u00a0<\/p>\n<ol>\n<li>W = hc\n<\/li>\n<\/ol>\n<p>\u00a0\u00a0\u00a0\u00a0       \u03bb<strong><sub>o<\/sub><\/strong><br \/>\n\t\t\t\t\u00a0\u00a0\u00a0\u00a0\u03bb<strong><sub>o<\/sub><\/strong> =hc<br \/>\n\u00a0\u00a0\u00a0\u00a0      W.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u03bb<strong><sub>o<\/sub><\/strong> =  6.6 x 10<sup>-34<\/sup> x 3.0 x 10<sup>8<\/sup><br \/>\n\t\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2.3 x 1.6 x 10<sup>-19<\/sup><br \/>\n\t\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u03bb<strong><sub>o<\/sub><\/strong> = 8.61 x 10<sup>-7<\/sup>m<\/p>\n<p>\u00a0<strong>EVALUATION<\/strong><\/p>\n<ol>\n<li>If photon of wave length 2.0 X 10<sup>17<\/sup>m is incident on a metal and the kinetic energy of the emitted electrons is 23.5eV . Calculate the work function of the metal. (h = 6.6 x 10<sup>-34<\/sup>JS, 1eV = 1.6 x 10<sup>-19<\/sup>J, c = 3.0 x 10<sup>8<\/sup> ).\n<\/li>\n<li>Determine the threshold frequency of the metal in (1) above, hence explain what will happen if a light of frequency 9.1 x 10<sup>22<\/sup><em>Hz<\/em> is illuminated on the metal.\n<\/li>\n<\/ol>\n<p>\u00a0<strong>X RAY<br \/>\n<\/strong>X-ray was discovered in 1895 by Williams Rontgen. X \u2013 rays are produced when thermally generated electrons from a hot filament are accelerated through a high potential difference and focused on to a tungsten target, where the electrons are suddenly stopped.<\/p>\n<p>\u00a0<img decoding=\"async\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0936_Week7SS3Se2.jpg\" alt=\"\"\/><\/p>\n<p>\u00a0<strong>MODE OF OPERATION<br \/>\n<\/strong>In the X- ray tube, a high potential difference is applied between the hot cathode and the anode. Electrons are emitted from the cathode and are accelerated to an extremely high speed. They are abruptly decelerated when they strike the anode causing the emission of high energy radiation of short wavelength i.e X-rays. The anode becomes very hot in the process and requires cooling gins on the outside of the tube.<\/p>\n<p>\u00a0<strong>ENERGY CONVERSION DURING X \u2013 RAY PRODUCTION<br \/>\n<\/strong>During X \u2013 ray production, electrical energy is converted to thermal energy. The thermal energy is converted into mechanical energy (kinetic energy) to accelerate the electron.  The mechanical energy is converted into electromagnetic energy of the x-ray <\/p>\n<p>\u00a0<strong>TYPES OF X \u2013 RAY<br \/>\n<\/strong>There are two types of x- rays<\/p>\n<ol>\n<li>Hard x \u2013 rays  and\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02. Soft x- rays\n<\/li>\n<\/ol>\n<p>\u00a0<strong>Characteristics of Hard x-rays<br \/>\n<\/strong><\/p>\n<ol>\n<li>High penetrating power or ability\n<\/li>\n<li>\n<div>Shorter wavelength\n<\/div>\n<p>\u00a0<\/li>\n<\/ol>\n<p><strong>Characteristics of soft x \u2013ray.<br \/>\n<\/strong><\/p>\n<ol>\n<li>low penetrating power\n<\/li>\n<li>\n<div>longer wavelength\n<\/div>\n<p>\u00a0<\/li>\n<\/ol>\n<p><strong>EVALUATION<\/strong><\/p>\n<ol>\n<li>state the energy conversions in an x-ray tube.\n<\/li>\n<li>\n<div>Differentiate between soft and hard x-ray.\n<\/div>\n<p>\u00a0<\/li>\n<\/ol>\n<p><strong>HARDNESS<br \/>\n<\/strong>This is a measure of the strength or penetrating ability of the x \u2013 ray.<\/p>\n<p>\u00a0<strong>INTENSITY<br \/>\n<\/strong>This is the energy radiated per unit time per unit area by the x \u2013ray. It depends on the current of the filament .<\/p>\n<p>\u00a0<strong>Properties of x- rays<br \/>\n<\/strong><\/p>\n<ol>\n<li>X \u2013 rays are electromagnetic waves of high frequency\n<\/li>\n<li>X \u2013 rays have short wavelength ( 2 x 10-10m )\n<\/li>\n<li>X \u2013 rays have high penetrating power\n<\/li>\n<li>X-rays travels in straight line\n<\/li>\n<li>They are not diffracted by electric or magnetic field.\n<\/li>\n<li>They are not diffracted by crystals.\n<\/li>\n<li>They ionized gases\n<\/li>\n<li>They cause zinc sulphide to fluoresce.\n<\/li>\n<\/ol>\n<p>\u00a0<strong>Application of X \u2013 ray<br \/>\n<\/strong><\/p>\n<ol>\n<li>For examining body to locate broken bones\n<\/li>\n<li>To detect metals and contra band in a baggage\n<\/li>\n<li>They are used to detect cracks n welded joints\n<\/li>\n<li>For investigating crystal structure\n<\/li>\n<li>Treatment of tumors and malignant growth\n<\/li>\n<li>It is used in agriculture to kill germs.\n<\/li>\n<\/ol>\n<p>\u00a0<strong>Hazards of x- rays<br \/>\n<\/strong><\/p>\n<ol>\n<li>It causes genetic mutation\n<\/li>\n<li>It can destroy body cells\n<\/li>\n<li>it causes leukemia, by damaging body tissues\n<\/li>\n<li>it causes skin burns and cancer.\n<\/li>\n<\/ol>\n<p>\u00a0<strong>Precautions<br \/>\n<\/strong>Those who work with x-rays should put on lead coat and they should always go for regular medical check-up.<\/p>\n<p>\u00a0<strong>GENERAL EVALUATION<\/strong><\/p>\n<ol>\n<li>What is the function of lighting conductors\n<\/li>\n<li>Explain lighting\n<\/li>\n<\/ol>\n<p>\u00a0<strong>WEEKEND ASSIGNMENT <\/strong><br \/>\n\t\t\t1.\u00a0\u00a0\u00a0\u00a0Which of the following give rise to the line spectra observed in atoms? (a) excitation of electrons in the atom (b) change of an electron from a higher to a lower energy level<br \/>\n(c) Distributed photo in the nucleus<br \/>\n2.\u00a0\u00a0\u00a0\u00a0Which of the following is called photo electric effect. (a) two electrons are created from a quantum of light (b) metals absorbs quanta of light and then emits electrons (c) a high  energy<br \/>\nemits  photon as it is slowed down<br \/>\n3.\u00a0\u00a0\u00a0\u00a0The minimum frequency that can cause photo emission of electrons from metal surface is known as (a) wavelength (b) threshold   frequency (c) frequency of the incident light<br \/>\n4.\u00a0\u00a0\u00a0\u00a0The maximum kinetic of the photo electrons depend on (a) work function (b) frequency<br \/>\n(c) intensity of the incident ray<br \/>\n5.\u00a0\u00a0\u00a0\u00a0The minimum energy required to liberate an electron from a metallic surface is (a) ionization energy\u00a0\u00a0\u00a0\u00a0(b) work function (c) kinetic energy,<\/p>\n<p>\u00a0<strong>THEORY<\/strong><br \/>\n\t\t\t1.\u00a0\u00a0\u00a0\u00a0(a) explain the terms &#8221; hardness&#8221; and &#8221; intensity&#8221; as applied to x-ray tube.<br \/>\n\u00a0\u00a0\u00a0\u00a0(b) State three uses of x-rays<\/p>\n<ol>\n<li>Determine the frequency of the photon whose energy is required to eject a surface electron with a kinetic energy of 1.970 x 10<sup>-19<\/sup> eV.  If the work function of the metal is 1.334 x 10<sup>-19<\/sup>eV.(1eV = 1.6 x 10<sup>-19<\/sup>J, h = 6.6 x 10<sup>-34<\/sup>Js, C = 3 .0 x 10<sup>8<\/sup>ms-<sup>1<\/sup>)\n<\/li>\n<\/ol>\n<p>\u00a0<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0WEEK SEVEN TOPIC: PHOTO ELECTRIC EFFECT CONTENT Application Photoelectric Equation Threshold frequency Work funtion x-ray&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,317],"tags":[],"class_list":["post-4097","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss3-physics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/4097","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=4097"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/4097\/revisions"}],"predecessor-version":[{"id":4098,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/4097\/revisions\/4098"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=4097"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=4097"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=4097"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}