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andreev551 [17]
3 years ago
6

Compare the de Broglie wavelength of an alpha particle moving at 3.40×107 miles per hour (1.52×107 m/s) to that of a baseball mo

ving at 90.0 miles per hour (40.2 m/s) and an electron with a speed of 1.30×107 miles per hour (5.81×106 m/s). Which region of the electromagnetic spectrum are each of these wavelengths near?
Chemistry
1 answer:
denis-greek [22]3 years ago
3 0

Answer:

1.74 x 10⁻²⁰(baseball) : 1.90 x 10⁴ (electron) : 1 (alpha particle)

baseball not detected

electron x ray

alpha particle beyond gama rays

Explanation:

The strategy for us here is to utilize deBroglie relation of wavelength and momentum:

λ = h/ mv

where  λ is wavelength,  h isPlanck´s constant, m is the mass of the particle, and v its velocity.

For determining the regions of spectrum the wavelengths fall into, we need to consult charts that describe the wavelengths vs region.

So  using the above  equation we will compute the wavelengths for the alpha particle, baseball,  and an electron.

Notice we are missing the masses of the alpha particle, and that of the electron, and baseball

The mass of the alpha particle is 4 times  amu  (4 x 1.66 x 10⁻²⁷ Kg ) since the alpha particle is esentially a helium nucleus which has atomic number 4.

The mass of the electron is 9.11 x 10⁻³¹ kg.

Mass of baseball per rules around 0.145 kg.

Notice we are working in the metric system, so use the velocities in m/s.

Now that we have all the data required, lets proceed to calculate the respective wavelengths.

λ ( alpha particle ) = 6.626 x 10⁻³⁴ J·s / ( 4 x 1.66 x 10⁻²⁷ kg x 1.52 x 10⁷ m/s )

= 6.57 x 10⁻¹⁵ m

λ ( baseball ) = 6.626 x 10⁻³⁴ J·s / ( 0.145 kg  x 40.2 m/s ) = 1.14 x 10⁻³⁴ m

λ ( electron ) = 6.626 x 10⁻³⁴ J·s / ( 9.11 x 10⁻³¹ kg x 5.81 x 10⁶ m/s )

= 1.25 x 10⁻¹⁰ m

Comparing the wavelengths from largest to smallest we have:

λ ( baseball) : λ ( electron )   : λ ( alpha particle )

1.14 x 10⁻³⁴  : 1.25 x 10⁻¹⁰ :  6.57 x 10⁻¹⁵

1.74 x 10⁻²⁰(baseball) : 1.90 x 10⁴ (electron) : 1 (alpha particle)

We can see the wavelength  of the baseball is very, very small compared to that of an electron and an alpha particle. For this small wavelength  we are not going to see effects such as diffraction or interference.This is the reason that for everyday macroscopic objects we do not talk about its associated wavelength. The wavelength can only be appreciated in  very small microscopic particles as exemplified for the cases of the electron and alpha particle in this question.

The wavelength of the baseball cannot be detected so a placement in the electromagnetic spectrum is undefined.

The wavelength of the electron in this question will fall into the x ray region of the spectrum (   region  around 10⁻⁹ to 10⁻¹² ) and the alpha particle will fall beyond the gamma rays ( that is wavelength shorter than 10⁻¹² m ).

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Explanation:

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Use the born-haber cycle to calculate the lattice energy of kcl. (δhsub for potassium is 89.0 kj/mol, ie1 for potassium is 419 k
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Given data:

Sublimation of K

K(s) ↔ K(g)                            ΔH(sub) = 89.0 kj/mol

Ionization energy for K

K(s) → K⁺ + e⁻                         IE(K) = 419 Kj/mol

Electron affinity for Cl

Cl(g) + e⁻ → Cl⁻                      EA(Cl) = -349 kj/mol

Bond energy for Cl₂

1/2Cl₂ (g) → Cl                        Bond energy = 243/2 = 121.5 kj/mol

Formation of KCl

K(s) + 1/2Cl₂(g) → KCl(s)        ΔHf = -436.5 kJ/mol

<u>To determine:</u>

Lattice energy of KCl

K⁺(g) + Cl⁻(g) → KCl (s)                   U(KCl) = ?

<u>Explanation:</u>

The enthalpy of formation of KCl can be expressed in terms of the sum of all the above processes, i.e.

ΔHf(KCl) = U(KCl) + ΔH(sub) + IE(K) + 1/2 BE(Cl₂) + EA(Cl)

therefore:

U(KCl) = ΔHf(KCl) - [ΔH(sub) + IE(K) + 1/2 BE(Cl₂) + EA(Cl)]

         = -436.5 - [89 + 419 + 243/2 -349] = -717 kJ/mol

Ans: the lattice energy of KCl = -717 kj/mol



5 0
3 years ago
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