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nyushanushash
@nyushanushash
September 2021
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определите энергетический выход ядерной реакции 3Li7 + 1H2 → 4Be9 + 0n1.
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belozerova1955
₃Li⁷+₁H²⇒₄Be⁹+₀n¹
Δm = (7,01601а.е.м.+2,01410а.е.м) - (9,01504а.е.м+1,00866а.е.м) = -0,99359а.е.м < 0 (реакция идет с поглощением энергии)
Е = 0,99359а.е.м.·931МэВ =925МэВ
2 votes
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Answers & Comments
Δm = (7,01601а.е.м.+2,01410а.е.м) - (9,01504а.е.м+1,00866а.е.м) = -0,99359а.е.м < 0 (реакция идет с поглощением энергии)
Е = 0,99359а.е.м.·931МэВ =925МэВ