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3 years 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:

Verizon [17]3 years ago

7 0

Answer:

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

Explanation:

From the question we are told 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 electron is $m_e = 9.109*10^{-31} \ kg$

Generally the energy of the incident light is mathematically represented as

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

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

h is the Planck constant with value $h = 6.62607015 * 10^{-34 } J\cdot s$

So

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

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

Generally the kinetic energy is mathematically represented as

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

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

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

Generally the ionization energy is mathematically represented as

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

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

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Answer:

Part(A): The magnitude of Jill's final velocity is $\bf{6.59~m/s}$ .

Part(B): The direction is $\bf{42.7^{0}}$ south to east.

Explanation:

Given:

Mass of Jack, $m_{1} = 59.0~Kg$

Mass of Jill, $m_{2} = 47..0~Kg$

Initial velocity of Jack, $v_{1i} = 8.00~m/s$

Initial velocity of Jill, $v_{2i} = 0$

Final velocity of Jack, $v_{1f} 5.00~m/s$

The final angle made by Jack after collision, $\alpha = 34.0^{0}$

Consider that the final velocity of Jill be $v_{2f}$ and it makes an angle of $\beta$ with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

$~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}$

where, $v_{2f}^{x}$ is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

$v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s$

Conservation of momentum of the system along north direction is given by

$~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s$

where, $v_{2f}^{y}$ is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

$v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s$

Part(B):

The direction is given by

$\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}$

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3 years ago