james-p

to try and replicate the cold fusion experiment by pons and fleishmen to see if its real

ok thanks

soo i am looking for some pure paladium a piece about 3 cm long and 1 cm wide and thickness does not matter. any suggestion's on wher i cold find some? thnx

so it sould be 3.055 x 10^-29 kg x 2997 x 10^5 = 9.155^-21

alredy did,all digits after the last number are ussless

thanks i finally figures it out is this right E=MC^2 = 0.00000000000000000000000000003055kg x (299792458)^2m/s =0.00000000000000000000000000003055kg x 89875517900000000 m/s =0.000000000002745697071845 J And 0.00000000000016021765 J equals to 1 MeV so 0.000000000002745697071845 J / 0.00000000000016021765 J equals to 17.137294622939482635028038421485 MeV.

soo how do you solve the equation when i put the number in it tells me their is only 0.000000000002745697071845 joules and not 18 MeV

ok so if i have a mass of 0.00000000000000000000000000003055kg how can i convert this into MeV energy using E=MC2 could you show me the calcules becaus idont think i did it the right way thanks

but on the graph it shows that deutirium is 1.2 mev and not 2 and tritium 2.9 mev not 8 did i solv the equation right if not could you show me how to thnx 2.0136 + 3.0160492 = 4.002602 + 1.008647668 D T He-4 5.0296492 = 5.011249668 D+T He-4 + N 5.0296492 - 5.011249668 = 0.018399532 D+T He-4 + N E=MC^2 =0.00000000000000000000000000003055kg x (3x10^8ms^-1)^2 D T c^2 =0.00000000000000000000000000003055kg x (3x 10^8ms^-1) x (3x10^8ms^-1) =0.00000000000000000000000000003055kg x (9x10^16m^2s^-2) =0.00000000000000000000000000003055kg x (9x10^16) kg m^2 s^-2 = 0.00000000000000000000000000003055kg x 90000000000000000 kg m^2 s^-2 =0.0000000000027495MeV

but wher does he get 2MeV 8MeV and 28MeV thanks

to calculate the energy released from a deutiriume tritium fusion to creat helium4 with a neutron do take the MeV from deutirium and add it to tritium the substact the MeV from He4 and the total left would be the energy your gaining? ex H2 MeV is 1.2 + H3 MeV is 2.9 =4.1 and He4 MeV is 7.1 so 7.1 - 4.1 = 3 soo 3 is he amount of energy your liberating? is this right or is their a difrent formula? and in this link how does he get difrent binding energy compared to the graphic? thanks james

thank you for your answer.

no it is not for homwork i have just heard from my friend that it was possible but my physics teacher told my class that it was impossible. although could you show me the equations if they are the same as my freinds. he showed me that he startes with the speed of transition = frequency of the emited photon x the wave lenght

is it possible to calculate plankes constant from only knowing the frequency of a emited phonton and the speed of transition. if so how thanks

i am in 11 grade and my science fair project is mostly reserch but i would like a small fusion expeiment to do also i just researeshed one about puting a tungsten cathode inside a duteriume solution and passing a electrical current through it. what is your opinion on this. http://www.amasci.com/weird/anode.txt thanks