What evidence from j. j. thomson's experiments with the cathode ray led to the plum pudding model of the atom?

I found the choices:

A. The beam was attracted to a positive charge and repelled by a negative charge, showing that atoms contain small, negatively charged particles.

B. The beam was attracted to a negative charge and repelled by a positive charge, showing that atoms contain small, positively charged particles.

C. The beam was positively charged and had significant mass, showing that atoms have a positive, dense nucleus.

D. The beam was neutral and had significant mass, showing that atoms contain neutral particles in the nucleus.

A. The beam was attracted to a positive charge and repelled by a negative charge, showing that atoms contain small, negatively charged particles.

Explanation:

J.J Thomson in 1897 carried out experiments on the gas discharge tube which led to the discovery of cathode rays also known as electrons. He observed the following properties:

The last property led the plum pudding model of the atom. In this model Thomson suggested an atom as pool of electron surrounded by positive charges based on the fact that cathode rays attracts positive charges.

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years