Pair production can only occur if the energy of the photon is bigger than the rest mass (E0 = m0*c^2) of electron and positron, because this is the energy needed to create these particle (conservation of energy). Excess energy will be kinetic energy of the electron and positron. Rest mass of electron and positron is 0.511 MeV each, so 1.022 MeV in total.
1000 meters = 1kmTo get from meters to km you need to divide by 1000.So 1022 meters is the same as 1.022 km.
one MeV has 1,000,000 eVs So one eV has 0.000001 MeVs
To convert joules to megaelectronvolts (MeV), you can use the conversion factor 1 MeV = 1.60218 x 10^-13 joules. Simply divide the energy in joules by this conversion factor to get the equivalent energy in MeV.
1022 add 1028 = 2050
To convert from mega electronvolts (MeV) to joules, you can use the conversion factor 1 MeV = 1.602 x 10^-13 joules. So, to convert 474 MeV to joules: 474 MeV * 1.602 x 10^-13 joules = 7.595 x 10^-11 joules.
The conversion factor from kelvin to mega-electronvolts (MeV) is approximately 8.617 x 10-5 MeV per kelvin.
1 MeV is equal to 1.6 x 10^-13 volts. Therefore, 5 MeV is equal to 8 x 10^-13 volts.
78.6154
None. Hydrogen and Helium are base elements.Helium is helium.Chemically yes, but with Nuclear Physics in stars:H + H --> D + e+ + v + 0.42 MeV, H + H --> D + e+ + v + 0.42 MeV, D + D --> He3 + n + 3.27 MeV (50%)H + H --> D + e+ + v + 0.42 MeV, H + H --> D + e+ + v + 0.42 MeV, D + D --> He3 + H + 4.03 MeV (50%)H + H --> D + e+ + v + 0.42 MeV, H + D --> He3 + gamma photon + 5.49 MeV, H + H --> D + e+ + v + 0.42 MeV, H + D --> He3 + gamma photon + 5.49 MeV, He3 + He3 --> He + H + H + 12.86 MeV (100%)H + D --> He3 + gamma photon + 5.49 MeV (100%)D + D --> He3 + n + 3.27 MeV (50%)D + D --> He3 + H + 4.03 MeV (50%)D + T --> He + n + 17.59 MeV (100%)T + T --> He + n + n + 11.33 MeV (100%)Four or two, depending on the kind of hydrogen isotope you are burning. As you can see in the equations above burning four ordinary atoms of hydrogen to ordinary helium is a complicated and slow process compared to burning two atoms of various isotopes of hydrogen, however a star has such tiny quantities of deuterium and tritium that depending on them for fusion helps little.
200 MeV is a tiny quantity, it may sound a lot but is actually about 3 x 10-11 Joules. The point is that in a large fission reactor there are billions of fissions occurring every second and each one releases 200 MeV. As 1 Joule = 1 Watt.sec, and 1 KWh = 3.6 x 106 Watt.secs, you can see that the question is rather unbalanced!
To convert the decimal number 1022 to base 3, you repeatedly divide the number by 3 and keep track of the remainders. Performing this process, 1022 in base 3 is represented as 1102212. Thus, 1022 in base 3 is written as 1102212.
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