Respuesta :
Answer:
a) m = 993 g
b) E = 6.50 × 10¹⁴ J
Explanation:
atomic mass of hydrogen = 1.00794
4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176
we know atomic mass of helium = 4.002602
difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158
fraction of mass lost = [tex]\dfrac{0.029158}{4.03176}[/tex]= 0.00723
loss of mass for 1000 g = 1000 × 0.00723 = 7.23
a) mass of helium produced = 1000-7.23 = 993 g (approx.)
b) energy released in the process
E = m c²
E = 0.00723 × (3× 10⁸)²
E = 6.50 × 10¹⁴ J
Answer:
(a) 992.87 g
(b) [tex]6.419\times 10^{14} J[/tex]
Solution:
As per the question:
Mass of Hydrogen converted to Helium, M = 1 kg = 1000 g
(a) To calculate mass of He produced:
We know that:
Atomic mass of hydrogen is 1.00784 u
Also,
4 Hydrogen atoms constitutes 1 Helium atom
Mass of Helium formed after conversion:
[tex]4\times 1.00784 = 4.03136 u[/tex]
Also, we know that:
Atomic mass of Helium is 4.002602 u
The loss of mass during conversion is:
4.03136 - 4.002602 = 0.028758 u
Now,
Fraction of lost mass, M' = [tex]\frac{0.028758}{4.03136} = 0.007133 u[/tex]
Now,
For the loss of mass of 1000g = [tex]0.007133\times 1000[/tex] = 7.133 g
Mass of He produced in the process:
[tex]M_{He} = 1000 - 7.133 = 992.87 g[/tex]
(b) To calculate the amount of energy released:
We use Eintein' relation of mass-enegy equivalence:
[tex]E = M'c^{2}[/tex]
[tex]E = 0.007133\times (3\times 10^{8})^{2} = 6.419\times 10^{14} J[/tex]
