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Alex73 [517]
3 years ago
8

Nuclear power plants produce energy using fission. One common fuel, uranium-235, produces energy through the fission reaction 23

5 92U+10n→fission fragments+neutrons+3.20×10−11 J/atom 92235U+01n→fission fragments+neutrons+3.20×10−11 J/atom What mass of uranium-235 is needed to produce the same amount of energy as the fusion reaction in Part A? Express your answer to three significant figures and include the appropriate units.
Chemistry
1 answer:
Vikentia [17]3 years ago
7 0

Answer:

mass of U-235  = 15.9 g (3 sig. figures)

Explanation:

1 atom can produce -------------------------> 3.20 x 10^-11 J energy

x atoms can produce ----------------------> 1.30 x 10^12 J energy

x = 1.30 x 10^12 / 3.20 x 10^-11

x = 4.06 x 10^22 atoms

1 mol ----------------------> 6.023 x 10^23 atoms

y mol ----------------------> 4.06 x 10^22 atoms

y = 0.0675 moles

mass of U-235 = 0.0675 x 235 = 15.8625

mass of U-235  = 15.9 g (3 sig. figures)

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Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th
lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

  • <em>c is the speed of light = 3 x 10⁸ m/s</em>
  • <em>f is the frequency of the wave</em>
  • <em>λ is the wavelength</em>

f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>

<em />

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

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