Question

Suppose 1 kg of Hydrogen is converted into Helium. a) What is the mass of the He produced? b) How much energy is released in this process?

Answer 1

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 = $\dfrac{0.029158}{4.03176}$= 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 2

Answer:

(a) 992.87 g

(b) $6.419\times 10^{14} J$

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:

$4\times 1.00784 = 4.03136 u$

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' = $\frac{0.028758}{4.03136} = 0.007133 u$

Now,

For the loss of mass of 1000g = $0.007133\times 1000$ = 7.133 g

Mass of He produced in the process:

$M_{He} = 1000 - 7.133 = 992.87 g$

(b) To calculate the amount of energy released:

We use Eintein' relation of mass-enegy equivalence:

$E = M'c^{2}$

$E = 0.007133\times (3\times 10^{8})^{2} = 6.419\times 10^{14} J$