1.In the fusion reaction 12H+12H→24He+Q^2_1\text{H} + ^2_1\text{H} \to ^4_2\text{He} + Q12H+12H→24He+Q, QQQ is energy released. If ccc is the speed of light and mmm is the mass of each deuterium nucleus then the mass of helium nucleus formed isa.2m+Q/c22m + Q/c^22m+Q/c2b.Q/mc2Q/mc^2Q/mc2c.m+Q/c2m + Q/c^2m+Q/c2d.2m−Q/c22m - Q/c^22m−Q/c2Login to continueOnly logged in users canattempt or see the solution.