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vovangra [49]
1 year ago
15

The hydrogen emission spectrum is shown below. What is the energy of the

Chemistry
1 answer:
Leviafan [203]1 year ago
8 0

The energy of the 434 nm emission line is 4.58×10¯¹⁹ J (Option A)

<h3>Data obtained from the question </h3>

The following data were obtained from the question:

  • Wavelength (λ) = 434 nm = 434×10¯⁹ m
  • Planck's constant (h) = 6.626×10¯³⁴ Js
  • Speed of light (v) = 3×10⁸ m/s
  • Energy (E) =?

<h3>How to determine the energy </h3>

The energy of the 434 nm emission line can be obtained as follow:

E = hv / λ

E = (6.626×10¯³⁴ × 3×10⁸) / 434×10¯⁹ 9.58×10¹⁴

E = 4.58×10¯¹⁹ J

Thus, the energy of the 434 nm emission line is 4.58×10¯¹⁹ J

Learn more about energy:

brainly.com/question/10703928

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Answer:

The Avogadro's  number is N_A     =  6.02289 *10^{23}

Explanation:

From the question we are told that

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    The density of the metal is \rho =  5.30\ g/cm^3 = 5.30 * \frac{g}{cm^3}  * \frac{1*10^6}{1*10^3} = 5.30 *10^3kg/m^3

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Generally the volume of a unit cell is  

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substituting value

        V =  [5.02 *10^{-10}]^3

         V = 1.265*10^{-28}\ m^3  

From the question we are told that 68% of the unit cell is occupied by Ba atoms and that the structure is a metal which implies that the crystalline structure will be  (BCC),

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substituting value

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     N_A     =   \frac{ 0.1373}{ 4.301*10^{-29} *  5.3*10^{3}}

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