Why does nuclear fusion reaction yield more energy than Nuclear fission reaction?

Fission only produces more energy than it consumes in larger nuclei (eg: Uranium & Plutonium) which have around 240 nucleons.

Fusion only produces more energy than it consumes in small nuclei (in stars, Hydrogen & its isotopes fusing into Helium).

The energy released when four Hydrogen nuclei fuse in to a Helium nucleus is around 27MeV or about 7 MeV per nucleon.

For fission of U or P, energies released are around 200 MeV or so. The energy per event is greater in fission, but the energy per nucleon (fusion = about 7MeV/nucleon; fission = about 1 MeV/nucleon) is much greater in fusion.

Now lets look at fission. An example of fission is when a U-235 atom is split by a neutron into a Ba-144 and Krypton-89 atoms and 3 neutrons. The binding energy per nucleon for Uranium is about 7.6 MeV and for Barium around 8.3 MeV giving an increase in binding energy during fission of about 0.7MeV per nucleon or a total of 164.5MeV in total.

In a fusion reaction firstly two Hydrogens form a Deuteron, a positron and an electron neutrino. Then the Deuterium fuses with  another Hydrogen to form  He-3 and a photon. Finally two He-3's fuse forming a Helium nucleus and two hydrogen nuclei.

Considering the mass of four  protons/hydrogen nuclei and the mass of helium produced we get a mass difference of 24.69 MeV.

Conclusion: We can conclude that fusion reactions give out more energy per reaction. This is 0.7 MeV for fission and 6.2 MeV for fusion.  

I think you have cleared your doubt. Have a nice time ............bye.

Why heavy Nuclei U-238 do not spontaneously undergo fission?

To understand this, one has to realize that the surface area is considerably increased when one large liquid drop breaks into two small liquid drops. In case of U-238, about 6MeV is required to overcome the surface forces and get separated into small fragments before energy is released. This situation is described in terms of potential barrier through which fission fragments have to pass.

The quantum mechanical probability for such a barrier penetration is considered. Because of large masses involved, this probability is extremely small and for U-238 we get a half life of approx. 10^16 years, for decay of spontaneous fission.

Actually for U-238, gamma rays of energy approx 6 MeV are able to induce fission reaction. This reaction is called photo fission for which threshold energy is 6MeV approx. So for all practical purposes, U238 is stable against decay by spontaneous fission.          

What makes a steel stainless?

The addition of Chromium and Nickel to the iron creates a significant percentage of Chromium and Nickel items at the surface.These atoms form tenacious oxides that mask the surface and prevent oxidation of the iron. The process known as passivation for stainless steel, is a common means of improving this protective oxide layer through the use of oxidizing acids.The corrosion resistance of stainless steel depends on the Chromium. 

Why does manufacturers coat inner walls of BF3 counters with activated charcoal?

The charcoal coating is used for tubes filled with BF3 gas and also for He3 filled counters operated in high neutron fluxes. The activated charcoal serves to absorb electronegative gases that build up during nuclear irradiation.

For instance, in a BF3 filled detector three fluorine atoms are released with each neutron capture.The Fluorine atoms will combine with electrons released in subsequent neutron captures. Initially, this process reduces electric pulse amplitude and eventually output pulses are eliminated altogether.