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Alika [10]
2 years ago
15

What is the de Broglie wavelength (in nm) of a proton of traveling at 3.18% of the speed of light?

Chemistry
1 answer:
gulaghasi [49]2 years ago
7 0
Answer is: <span>de Broglie wavelength of a proton is </span>3,4·10⁻⁵ nm.
v(proton) = 0,038 · 3·10⁸ m/s.
v(proton) = 1,14·10⁷ m/s; speed of proton.
m(proton) = 1,67·10⁻²⁷ kg.
h = 6,62607004·10⁻³⁴ m²·kg/s; Planck constant.
λ(proton) = h / m(proton) · v(proton).
λ(proton) = 6,62607004·10⁻³⁴ m²·kg/s ÷ (1,67·10⁻²⁷ kg · 1,14·10⁷ m/s).
λ(proton) = 3,48·10⁻¹⁴ m · 10⁹ nm/m = 3,4·10⁻⁵ nm.
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Convert the following temperatures to degrees Celsius or Fahrenheit: (a) 95°F, the temperature on a hot summer day; (b) 12°F,
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Answer:

a) 35°C, the temperature on a hot summer day

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Explanation:

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a) 95°F ⇒ °C = ( 95 - 32 ) / 1.8 = 35°C

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c) 102°F ⇒ °C = ( 102 -32 ) / 1.8 = 38.88°C

d) 1852°F ⇒ °C = ( 1852 -32 ) / 1.8 = 1011.11°C

e) 273.15°C ⇒ °F = ( 1.8 * 273.15 ) + 32 = 523.67°F

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