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zhannawk [14.2K]
3 years ago
6

The work function for metallic potassium is 2.3 eV. Calculate the velocity in km/s for electrons ejected from a metallic potassi

um surface by light of wavelength 260 nm.
Chemistry
1 answer:
mote1985 [20]3 years ago
5 0

Answer:

932.44 km/s

Explanation:

Given that:

The work function of the magnesium = 2.3 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, work function = 2.3\times 1.6022\times 10^{-19}\ J=3.68506\times 10^{-19}\ J

Using the equation for photoelectric effect as:

E=\psi _0+\frac {1}{2}\times m\times v^2

Also, E=\frac {h\times c}{\lambda}

Applying the equation as:

\frac {h\times c}{\lambda}=\psi _0+\frac {1}{2}\times m\times v^2

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

m is the mass of electron having value 9.11\times 10^{-31}\ kg

\lambda is the wavelength of the light being bombarded

\psi _0=Work\ function

v is the velocity of electron

Given, \lambda=260\ nm=260\times 10^{-9}\ m

Thus, applying values as:

\frac{6.626\times 10^{-34}\times 3\times 10^8}{260\times 10^{-9}}=3.68506\times 10^{-19}+\frac{1}{2}\times 9.11\times 10^{-31}\times v^2

3.68506\times \:10^{-19}+\frac{1}{2}\times \:9.11\times \:10^{-31}v^2=\frac{6.626\times \:10^{-34}\times \:3\times \:10^8}{260\times \:10^{-9}}

\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=7.64538\times 10^{-19}-3.68506\times 10^{-19}

\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=3.96032462\times 10^{-19}

v^2=0.869446\times 10^{12}

v = 9.3244 × 10⁵ m/s

Also, 1 m = 0.001 km

<u>So, v = 932.44 km/s</u>

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2) A common "rule of thumb" -- for many reactions around room temperature is that the
babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

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If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

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So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$  and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}} $

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$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

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$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}=  3.479 \times 10^{-6}$

$\Rightarrow  \frac{K_2}{K_1}=  e^{3.479 \times 10^{-6}}$

$\Rightarrow  \frac{K_2}{K_1}=  1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

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How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?
aksik [14]

Complete Question:

To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride. How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?

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2.23x10⁶ g

Explanation:

The concentration of the fluoride (F⁻) must be 0.800 ppm, which is 0.800 parts per million, so the water must have 0.800 g of F⁻/ 1000000 g of the solution. The density of the water at room temperature is 997 kg/m³ = 997x10³ g/m³. So, the concentration of the fluoride will be:

0.800 g of F⁻/ 1000000 g of the solution * 997x10³ g/m³

0.7976 g/m³

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A = πR², where R is the radius = 1.01x10² m (half of the diameter)

A = π*(1.01x10²)²

A = 32047 m²

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m = 0.7976 * 2.7983x10⁶

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Hence, option B is correct

Kindly Mark Brainliest, Thanks!!!

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