J.J Thomson's Experiment:
J.J Thomson devised an instrument to measure the e/m value of an electron using a discharge tube, electric, and
magnetic fields. By adjusting the strengths of these fields, he determined the e/m value to be 1.7588 x
1011
coulombs per kilogram.
Millikan's Oil Drop Method:
In 1909, Millikan determined the charge on an electron using his oil drop experiment. By observing the motion of
charged oil droplets in an electric field, he calculated the charge on a single electron to be 1.6022 x
10-19
coulombs, which is very close to the modern value.
What is the significance of measuring the e/m value of an electron?
The e/m value of an electron is crucial as it helps determine the charge-to-mass ratio of an electron, which is essential in understanding the fundamental properties of electrons and their behavior in electric and magnetic fields.
How did J.J. Thomson measure the e/m value of an electron?
J.J. Thomson measured the e/m value of an electron by using a discharge tube apparatus where cathode rays were passed through electric and magnetic fields. By adjusting the strengths of these fields, he was able to determine the e/m value of electrons, which was found to be 1.7588 x 10^11 coulombs per kilogram.
Why is the e/m value of positive rays different for different gases, while for cathode rays, the e/m values are the same?
The e/m value of positive rays varies for different gases because the positive rays consist of positively charged particles that have different masses. On the other hand, cathode rays are composed of electrons, which have the same mass, leading to consistent e/m values for cathode rays.
Why is the e/m value for positive rays obtained from hydrogen gas 1836 times less than that of cathode rays?
The e/m value for positive rays obtained from hydrogen gas is 1836 times less than that of cathode rays because positive rays are composed of positively charged particles that are much heavier than electrons. This difference in mass results in the significant variation in the e/m values between positive rays from hydrogen gas and cathode rays.
How can the mass of an electron be evaluated from the experiments conducted by Millikan and J.J. Thomson?
The mass of an electron can be evaluated by combining the charge of an electron obtained from Millikan's oil drop experiment (1.6022 x 10-19 coulombs) with the e/m value of an electron determined by J.J.Thomson (1.7588 x 1011 coulombs per kilogram). By using these values in the appropriate formula, the massof an electron can be calculated as 9.1095 x 10-31 kilograms.
What is the value of e/m for electrons according to J.J. Thomson's experiment?
a) 1.7588 x 1011 coulombs kg-1
b) 1.6022 x 10-19 coulombs
c) 9.1095 x 10-31 kg
d) 1.0073 amu
Answer: a) 1.7588 x 1011 coulombs kg-1
How did J.J. Thomson measure the e/m value of electrons?
a) Using a cathode ray tube
b) By observing the deflection of cathode rays in electric fields
c) By comparing electric and magnetic fields on cathode rays
d) Through the oil drop experiment
Answer: c) By comparing electric and magnetic fields on cathode rays
What happens to the atomic number when a radioactive Cu-65 captures a neutron?
a) It decreases by one unit
b) It remains the same
c) It increases by one unit
d) It becomes unstable
Answer: c) It increases by one unit
Why are cathode rays deflected in a circular path in the presence of a magnetic field?
a) Due to their positive charge
b) Due to their high speed
c) Due to their negative charge
d) Due to their mass
Answer: c) Due to their negative charge
What is the smallest charge found by Millikan in his oil drop experiment?
a) 1.7588 x 1011 coulombs kg-1
b) 1.6022 x 10-19 coulombs
c) 9.1095 x 10-31 kg
d) 1.0073 amu
Answer: b) 1.6022 x 10-19 coulombs