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6 days ago

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The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ei. This energy ca

n be measured by a technique called photoelectron spectroscopy, in which light of wavelength λ is directed at an atom, causing an electron to be ejected. The kinetic energy of the ejected electron (Ek) is measured by determining its velocity, υ (Ek= mυ2/2), and Ei is then calculated using the conservation of energy principle. That is, the energy of the incident light equals Ei plus Ek. What is the ionization energy of selenium atoms in kJ/mol if light with λ = 48.2 nm produces electrons with a velocity of 2.371x106 m/s? The mass, m, of an electron is 9.109x10-31 kg. (Round to the ones place.)

1 answer:

Keith_Richards [1K]6 days ago

8 0

Answer:

The resulting value is $E_i = 1.5596 *10^{-18} \ J$ .

Explanation:

The question specifies that

The wavelength is $\lambda = 48.2 nm = 48.2 *10^{- 9 }\ m$ .

The velocity is $v = 2.371*10^6 \ m/s$ .

The mass of the electron is $m_e = 9.109*10^{-31} \ kg$ .

The energy of the incoming light is typically depicted mathematically as

$E = \frac{h * c}{\lambda}$ .

Here, c represents the speed of light with the value $c = 3.0 *10^{8} \ m/s$ .

h stands for Planck's constant with a value of $h = 6.62607015 * 10^{-34 } J\cdot s$ .

Thus,

$E = \frac{6.62607015 * 10^{-34 }* 3.0 *10^{8}}{48.2 *10^{- 9 }}$

=> $E = 4.12 *10^{-18} \ J$ .

Typically, kinetic energy is represented as

$E_k = \frac{1}{2} * m_e * v^2$

=> $E_k = \frac{1}{2} * 9.109*10^{-31} * (2.371*10^6 )^2$ .

=> $E_k = 2.56 *0^{-18} \ J$ .

The ionization energy is generally expressed mathematically as

$E_i = 4.12 *10^{-18} - 2.56 *0^{-18}$

=> $E_i = 1.5596 *10^{-18} \ J$ .

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Answer:

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Explanation:

Given that,

Moles present = 3

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Ending temperature = 320 K

We are tasked with determining the heat absorbed by the gas at constant pressure

Employing the heat formula

$\Delta H_{1}=nC_{p}\times\Delta T$

Substituting the values into the equation

$\Delta H_{1}=3\times6.9\times(320-300)$

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Next, we calculate the heat absorbed by the gas at constant volume

Using the corresponding heat formula

$\Delta H_{1}=nC_{v}\times\Delta T$

Insert the values into the formula

$\Delta H_{1}=3\times4.9\times(300-320)$

$\Delta H_{1}=-294\ cal$

Now, it's necessary to evaluate the total heat flow into the gas during both steps

Using the total heat formula

$\Delta H_{T}=\Delta H_{1}+\Delta H_{2}$

$\Delta H_{T}=414-294$

$\Delta H_{T}=120\ cal$

Thus, the heat that transfers into the gas throughout this two-step process amounts to 120 cal.

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The equivalent distance in kilometers is 4012 × $10^{-6}$ km.

Explanation:

It's known that 1 millimeter converts to $10^{-3}$ meters. Then, 1 meter converts to $10^{-3}$ kilometers. Therefore, the conversion for 1 millimeter to kilometers can be stated as

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Thus, 1 mm = $10^{-3}$ × $10^{-3}$ km = $10^{-6}$ km.

Given the distance of 4012 mm, the corresponding distance in kilometers will be

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The distance therefore is 4012 × $10^{-6}$ km.

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Here, $\theta$ and $\theta _{0}$ denote the final and initial angular displacements, respectively, whereas $\omega$ and $\omega_{0}$ represent final and initial angular velocities, and $\alpha$ is the angular acceleration.

We are provided with $\alpha = - 25.60 \ rad/s^2, \theta = 62.4 \ rad$ and $t = 4.20 \ s$ .

By substituting these values into equation (A), we have

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Answer:

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For convenience, we will utilize a positive notation for downward acceleration and a negative notation for upward acceleration.

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The net force is expressed as:

-m(a/2) = mg - F (The balloon is moving upwards, hence F > mg)

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