J J Thomson and Millikan experiment

Question 5.1:What is J. J. Thomson’s experiment for determining e/m value of electron? Answer:The charge to mass ration(e/m) was determined by J.J. Thomson in 1897. The apparatus consists of a discharge tube shown in fig.

Figure: Measurement of e/m value of electron

The cathode rays can pass through electric and magnetic fields. Electric field is on the plane of paper and magnetic field is perpendicular to paper in the diagram. When both fields are off then beam of cathode rays produce luminous spot at point P1 on the screen. When only magnetic field is on then cathode rays strike at the point p3. When only electric field is applied, cathode rays strike at spot p2. Both electric and magnetic fields are then applied simultaneously, and their strength adjusted in such a way that cathode rays again hit point p1. In this way by comparing the strength of two fields we can determine e/m value of electron. It is 1.7588 x 1011coulombs kg-1.

Question 5.2: Explain Millikan’s oil drop experiment to determine the charge of an electron.
Answer: In 1909, Millikan determined the charge on the electron by oil drop method. The apparatus consists of a metallic chamber filled with air whose pressure can be adjusted by a vacuum pump. There are two electrodes; positively charged A with a hole in it and negatively charged A/. When a droplet of oil created by atomizer pass through the hole of upper plate A/ and fall between illuminated space between A and A/, its velocity (V1) is observed by microscope. Velocity v1 of the droplet of mass m depends upon its weight.


Figure: Millikan oil drop method for determination of charge on electron

After that the droplet takes electron from ionization of air by x-rays and gets charged. Now A and A/ are connected to battery and electric field of strength E is generated between these plates. The droplet moves upwards against the gravity with velocity V2.

Equation Where e is charge on the drop and Ee is the upward driving force due to electric field. Dividing equation (1) by equation (2) we get equation

The values of V1, and V2 are recorded with help of telescope and that of E and g are also known. Hence value of e can be calculated. The smallest value is 1.59 x10-19 which is very close to recent value of 1.6x 10-19. This smallest charge on any droplet is due to addition of single electron to the droplet. So, this is charge on electron. Other drops having more than one electron on them have double or triple amount of this charge.

Question 5.3: Evaluate mass of electron from charge to mass ration and Millikan experiment.
Answer: From Milliken’s experiment, value of charge on electron, e = 1.602 x 10-19 coulombs.

Next

Comments

Class 9 Chemistry Notes for Federal Board

Class 9 Chemistry Notes for Federal Board Introduction Class 9 Chemistry notes for the Federal Board have been written according to the National Curriculum 2006. It develops various skills such as questioning, observations, predicting, measuring, classifying, designing experiments, analyzing and interpreting the data. among students. We have made these notes very simple. After reading these notes, students can easily understand all topics. Students can answer all types of questions including essay type, short and multiple choice, self assessments and think tank. We have also solved exercises according to the Federal board examination pattern. Students, after going through 9th class chemistry notes can get maximum marks in their exams. They can lay down a strong foundation of scientific knowledge among students. In future, notes for class 9 will be available in pdf format. Students will be able to note pdf free download very soon. Students will have access to these notes in a mobi

Class 10 Chemistry Notes for Federal Board

Class 10 Chemistry Notes for Federal Board Introduction Class 10 Chemistry Notes for Federal Board have been written according to the National Curriculum 2006. These notes will make students able to learn topics by their own. 10th class notes for all chapters are very and easy to understand. After studying these notes, students will become confident to answer all type of questions. Chemistry notes for class 10 provide solution to all kinds of questions including essay type, short and multiple choice, self assessments and think tank. These notes also contain all past papers of reduced syllabus as well as complete syllabus. Soon, students will have access to 10th class chemistry notes in Urdu pdf download form. We are also working hard to provide all chemistry notes in pdf format and student will be able to download these notes. We also trying to upload video lectures about class 10 chemistry. Don't forget, you can test your chemistry by using online test. Furthermore, you highl

Theories of Covalent Bonding and Shapes of Molecules: MCQs

chemicalAlgebra 1. The bond energies of F 2 , Cl 2 , Br 2 and I 2 are 37, 58, 46 and 36 Kcal, respectively. The strongest bond is formed in A) F 2 B) Cl 2 C) Br 2 D) I 2 ... Answer is B) Description: The bond energy is proportional to bond length. So, the bond strength of Cl 2 is higher. show Answer 2. The percentage of ionic character of bond between two atoms is calculated from the difference between their A) Ionic B) Electronegativities C) Electron affinities D) Ionization energies ... Answer is B) Description: The percentage of ionic character depends upon the difference of E.N. difference between the bonded atom. Greater the E. N. difference, greater the ionic character and hence greater the bond energy show Answer 3. The geometry of PF 5 molecule is A) Planer B) Square planer C) Trigonal pyramidal D) Tetrahedral ... Answer is C) Description: As i

Chemistry

T he Federal Board of Intermediate & Secondary Education (FBISE) is an autonomous body that conducts the examinations in the institutions affiliated with it in Pakistan as well as abroad. Punjab government has ten boards of Intermediate & Secondary Education including Bahawalpur Board, D.G. Khan Board, Faisalabad Board, Gujranwala Board, Lahore Board, Multan Board, Rawalpindi Board and Sargodha Board and each board conducts exams independently. Federal Board generally follows the international norms and standards. Federal Board paper pattern is quite tricky and similar to O-Level and A Level formats. These questions need conceptual approach to be solved. Punjab board’s Paper pattern is comparatively easy up to some extent. We are currently focusing on the subject of Chemistry for all classes affiliated with Federal Board ( IX , X , XI , XII ) and Punjab Board ( XI XII ). We intend to extend it to all subjects in future. After reading All Student Notes, studen

Basic Definitions| Matter, Substance, Element, Compound. Mixture, Homogeneous mixture and Heterogeneous mixture Examples (for Grade - 9 students)

Basic Definitions| Matter, Substance, Element, Compound. Mixture, Homogeneous mixture and Heterogeneous mixture Examples (for Grade - 9 students) Question: Define following terms: Matter Substance Element Compound Mixture Homogeneous mixture Heterogeneous mixture Answer Matter: Matter can be anything that occupies space (volume) and has mass. Example, book, table, walls, chairs, plants, animal, etc. Substance: Substance is a matter that has a particular set of characteristics that differ from the characteristics of another kind of matter. For example, oxygen, water, carbon monoxide, carbon dioxide, urea, glucose, common salt etc are different substances. Element: A substance that cannot be converted to other simpler substances is known as an element. or A substance which has all atoms of same atomic number is called element. For example substances like oxygen, hydrogen, iron, copper, aluminum etc. are elements. Compound: A compound is a pure substance that co

Experimental Techniques in Chemistry

Analytical Chemistry: The branch of chemistry that deals with the quantitative and qualitative analysis of substances is called Analytical Chemistry. Question 2.1: What is difference between quantitative and quantitative analysis? What are various steps involved in quantitative analysis? Answer: Qualitative analysis: The analysis in which elements are identified is called qualitative analysis. Quantitative analysis: The analysis in the form of which relative amounts of elements are determined is called quantitative analysis. There are four main steps involved in quantitative analysis. 1. Obtaining a sample for analysis 2. Separation of desired constituent 3. Measurement and calculations of results 4. Drawing conclusions from analysis Question 2.2: Define filtration. How do we select the filter media? Answer: The process of separating solids or suspended particles from liquids by using a filter medium is called filtration. Question 2.3: Describe filt

Alkyl halides

The compounds in which one hydrogen atom of alkane is replaced by a halogen atom is called alkyl halide. Alternatively, The halogen derivatives of alkanes with general formula R-X are Halide called alkyl halide. For example CH4 ( Alkane ) CH3-X ( Alkyl halide ) X = F, Cl, Br and I Table Of Contents Types Mono, di, tri or poly haloalkane depending upon the number of halogen atom. For example C lassification of Alkyl Halides Alkyl halides can be classified as primary, secondary and Tertiary alkyl halides. 1. Primary Alkyl Halides The alkyl halides in which a halogen atom is bonded with primary carbon is called primary alkyl halide. Primary carbon is the carbon that is attached one or no carbon atom. The Secondary Alkyl Halides The alkyl halides in which a halogen atom is bonded with secondary carbon are called secondary alkyl halide. Secondary carbon is the carbon that is attached with two carbon atoms. For example

CHAPTER # 15 ORGANIC COMPOUNDS

ORGANIC COMPOUNDS Modern definition. C ompounds of carbon, hydrogen and their derivatives, are called organic compounds. For example, CH 4 , CH 3 Cl etc. Contents Sources of organic Compounds: Following are the main sources of organic compounds: i. Fossil fuels or Fossil remains. ii. Plants and Natural product iii. Partial &Total Synthesis iv. Fermentation and Biotechnology Fossil fuels or Fossil remains: The main sources of fossil fuels or fossil remains are: a. coal b. petroleum c. gas a. Coal: Coal is one of the organic compounds. Coal major source of into coke, and coal law" on pyrolysis Com can be converted destructive distillation. More than 200 organic compounds" have been directly isolated from coal tar, which are used. in ...drug. dyes, other. preparation of perfumes.. photographic developers and. b. Petroleum: Petroleum Sticky Liquid ground in that is block Thick seeps out of the some part of world. It is Complex mixtur

Shielding Effect (Grade 9 Students) and its trend in Periodic Table.

Question: What is shielding effect? Give its trends in the periodic table. Answer . Shielding Effect The decrease in force of attraction between nucleus and valence electrons due to presence of inner shell electrons is called shielding effect. It is also known as screening effect. It increases with increase in inner shell electrons. For example, Mg has greater shielding effect than Be. This is because Mg has eight more inner-shell electrons than Be. Trend in Periodic Table In period As we move from left to right in a period, shielding effect remains constant. Reason is that the number of inner shell electrons remains constant along the period. In group As we move from top to bottom in a group, shielding effect increases. This is due to fact that the number of inner shell increases and hence, inner shell electrons also increase. For example, Na has greater shielding effect than Li because of eight more inner shell electrons than Li. Example 3.5 Identi

Class 9 Electrochemistry MCQs and Answers with explanation

1. In which of the following changes, the nitrogen atom is reduced? a) N 2 to NO b) N 2 to NO 2 c) N 2 to NH 3 d) N 2 to HNO 3 ... Answer is b) Explanation: Oxidation number of an elemnet in free state is zero. So Oxidation No. of N 2 = 0 Calculate oxidation No. of nitrogen in each compound. Option a) Let, oxidation No. of N in NO is X. Sum of oxidation numbers of all atoms present in a neutral compound is zero. Oxidation No. of N + (Oxidation No. of O ) = 0 X-2 = 0 X = +2 X = +2 Oxidation No. of N in NO 2 is +2 Option b) Let, oxidation state of N in NO 2 is X. Sum of oxidation numbers of all atoms present in a neutral compound is zero. Oxidation No. of N + 2(Oxidation No. of O ) = 0 X + 2(-2) = 0 X -4 = 0 X = +4 Oxidation No. of N in NO 2 is +4 Option c) Let, oxidation state of N in NH 3 is X. Sum of oxidation numbers of all atoms present in a neutral compound is zero. Oxidation No. of N + 3(Oxidation No. of H )

All Students Notes

Search This Blog