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Tpy6a [65]
3 years ago
10

A laser with a wavelength of 225 nm is shown on an isolated gas-phase sodium atom. Calculate the velocity of the ejected electro

n from the ionized atom. The first ionisation energy of sodium is 496 kJ/mol and the mass of an electron is 9.109 x 10^-31 kg.
Chemistry
1 answer:
monitta3 years ago
8 0

Answer:

The velocity of the ejected electron from the ionized atom is 3.6 × 10⁵ m/s

Explanation:

Using the conservation of energy, we can write that

Photon energy (E) = Ionisation energy (I.E) + Kinetic energy (K.E)

Photon  energy, E = hf

Where h is Planck's constant ( h = 6.626 × 10⁻³⁴ kgm²/s)

and f is the frequency

Also,

Kinetic energy, K.E = \frac{1}{2} mv^{2}

Where m is mass

and v is velocity

Hence, we can write that

hf = I.E + \frac{1}{2}mv^{2}

But, c = f\lambda

where c is the speed of light ( c = 3.0 × 10⁸ m/s)

and λ is the wavelength

∴ f = \frac{c}{\lambda}

Then,

\frac{hc}{\lambda}  = I.E + \frac{1}{2}mv^{2}

From the question, the first ionisation energy of sodium is 496 kJ/mol

This is the ionisation energy for 1 mole of sodium,

For 1 atom of sodium, we will divide by Avogadro's constant

∴ The ionisation energy becomes

(496 KJ/mol) / (6.02 × 10²³ molecules)

= 8.239 × 10⁻¹⁹ J

This is the ionisation energy for one atom of sodium

Now, to determine the velocity of the ejected electron from the ionized atom,

From,

\frac{hc}{\lambda}  = I.E + \frac{1}{2}mv^{2}

Then,

\frac{6.626\times 10^{-34} \times 3.0 \times 10^{8}  }{225 \times 10^{-9} } = 8.239 \times 10^{-19} + \frac{1}{2}(9.109\times10^{-31} )v^{2}

8.835 \times 10^{-19} = 8.239 \times 10^{-19} + 4.5545 \times 10^{-31}v^{2}8.835 \times 10^{-19} - 8.239 \times 10^{-19}   = 4.5545 \times 10^{-31}v^{2}

5.96 \times  10^{-20}   = 4.5545 \times 10^{-31}v^{2}

v^{2} = \frac{5.96 \times  10^{-20}}{4.5545 \times 10^{-31}}

v^{2} = 1.3086 \times 10^{11}

v = \sqrt{1.3086 \times 10^{11} }

v = 361745.77 m/s

v = 3.6 \times 10^{5}  m/s

Hence, the velocity of the ejected electron from the ionized atom is 3.6 × 10⁵ m/s

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Explanation:

a. Silver iodate

Let s = the molar solubility.  

                     AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹:                               s               s

K_{sp} =\text{[Ag$^{+}$][IO$_{3}$$^{-}$]} = s\times s =  s^{2} = 3.0\times 10^{-8}\\s = \sqrt{3.0\times 10^{-8}} \text{ mol/L} = 1.7 \times 10^{-4} \text{ mol/L}

b. Barium sulfate

                     BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹:                                0.02             0

C/mol·L⁻¹:                                 +s              +s

E/mol·L⁻¹:                            0.02 + s          s

K_{sp} =\text{[Ba$^{2+}$][SO$_{4}$$^{2-}$]} = (0.02 + s) \times s \approx  0.02s = 1.1\times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{0.02} \text{ mol/L} = 5.5 \times 10^{-9} \text{ mol/L}

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is  

\mu = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\\mu = \dfrac{1}{2} (\text{[Ba$^{2+}$]}\cdot (2+)^{2} + \text{[NO$_{3}$$^{-}$]}\times(-1)^{2}) = \dfrac{1}{2} (\text{0.02}\times 4 + \text{0.04}\times1)= \dfrac{1}{2} (0.08 + 0.04)\\\\= \dfrac{1}{2} \times0.12 = 0.06

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a. Calculate the activity coefficients of the ions

\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(1)^{2}\sqrt{0.06} = -0.51\times 0.24 = -0.12\\\gamma = 10^{-0.12} = 0.75

b. Calculate the solubility

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K_{sp} =\text{[Ag$^{+}$]$\gamma_{Ag^{+}}$[IO$_{3}$$^{-}$]$\gamma_{IO_{3}^{-}}$} = s\times0.75\times s \times 0.75 =0.56s^{2}= 3.0 \times 10^{-8}\\s^{2} = \dfrac{3.0 \times 10^{-8}}{0.56} = 5.3 \times 10^{-8}\\\\s =2.3 \times 10^{-4}\text{ mol/L}

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a. Calculate the activity coefficients of the ions

\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(2)^{2}\sqrt{0.06} = -0.51\times16\times 0.24 = -0.50\\\gamma = 10^{-0.50} = 0.32

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

K_{sp} =\text{[Ba$^{2+}$]$\gamma_{ Ba^{2+}}$[SO$_{4}$$^{2-}$]$\gamma_{ SO_{4}^{2-}}$} = (0.02 + s) \times 0.32\times s\times 0.32 \approx  0.02\times0.10s\\2.0\times 10^{-3}s = 1.1 \times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{2.0 \times 10^{-3}} \text{ mol/L} = 5.5 \times 10^{-8} \text{ mol/L}

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