find the binding energy and binding energy per nucleon for 26 Fe ^56

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Solution

Dear Student,

$L e t g i v e n i s, m a s s o f n e u t r o n = 1.008665 a m u m a s s o f p r o t o n = 1.007825 a m u m a s s o f {}_{26}{F e}^{56} = 55.934939 a m u {}_{26}{F e}^{56} h a s 26 p r o t o n a n d (56 - 26 =) 30 n e u t r o n s M a s s o f 26 p r o t o n s a n d 30 n e u t r o n s = 26 \times 1.007825 + 30 \times 1.008665 = 56.46340 a m u ∆ m = 56.46340 - 55.934939 = 0.528461 a m u W e k n o w, 1 a m u c o n v e r t t o 931.5 M e V B i n d i n g e n e r g y = 0.528461 \times 931.5 = 492.26 M e V B i n d i n g e n e r g y p e r n u c l e o n = \frac{492.26}{56} = 8.790 M e V$

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Similar questions

_{26} F^{56}

_{26} F e^{56}

_{83} B i^{209}

= 1.007825

= 1.08665

_{26} F e^{56}

= 55.934939

_{83} B i^{209}

= 208.980388

_{8}^{17} O \to 131 M e V

_{26}^{56} F e \to 493 M e V

_{92}^{238} U \to 1804 M e V

(Δ = \frac{B}{A})

1._{8}^{17} O \to 131 M e V

2._{26}^{56} F e \to 493 M e V

3._{92}^{238} U \to 1804 M e V

(Δ = \frac{B}{A})

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