(an easy problem which will be graded). later in the quarter we will spend some time solving the diffusion equation op(r, t) = dv4p(r, t), at where d is called the diffusion constant. for now, let us consider a solution which represents a steady state solution (i. e apat = 0) and which has spherical symmetry. in that case the above reduces to 1d (20p) = 0 p2 dr dr / a) obtain the solution of this second-order differential equation. it will have two arbitrary constants. b) application: diffusion from a large distance to a spherical absorber. consider p(r) to represent the density of some particles which are diffusing through some media. think of them as food for a cell. their density at large distance (infinity) is po. there is a spherical absorber of radius r centered at the origin, which means p(r) = 0. (this absorber represents the cell.) show that the particular solution that satisfies these two condition is p(r) = po(1 – r/r). an interesting result: the current density (the number of particles passing through unit area per unit time) is given by j(r) = -ddp/dr = -dpor/r2. (the minus sign means that the flow is inward.) the magnitude of the current at a distance r is i = 47r2|j(r)= 47 dpor. note that this result says that the rate at which particles are absorbed is proportional to the radius of the sphere, not to the area of the sphere as would be the case if we were dealing with light being absorbed. of course light does not travel by diffusion!

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases . this process does not change the mass number.

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as -decay. in this the mass number remains same.

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

what experimental evidence led scientists to change from the previous model to this one?

a few of the positive particles aimed at a gold foil seemed to bounce back.

the colors of light emitted from heated atoms had very specific energies.

experiments with water vapor showed that elements combine in specific proportions.

cathode rays were bent in the same way whenever a magnet was brought near them.