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yarga [219]
2 years ago
13

Explain why it takes more energy to remove the second electron from a lithium atom than it does to remove the fourth electron fr

om a carbon atom.
Physics
2 answers:
slava [35]2 years ago
8 0
It takes more energy to remove the second electron from a lithium atom than it does to remove the fourth electron from a carbon atom because its inner core e, not valence e. C's 4th removed e is still a valence e. And also <span>because more nuclear charge acting on the second electron, it is more close to the nucleus, thus the the protons attract it more than the 4th electron.</span>
Orlov [11]2 years ago
8 0

The energy required to remove second electron from lithium is more as compared to removing fourth electron from carbon

EXPLANATION:

The amount of energy required to remove an electron from an isolated atom is called as ionization energy of the electron.

The second ionization energy of lithium atom is more as compared to fourth ionization energy of carbon atom. It is so because the second electron which is to be emitted from the K-shell of lithium atom, is tightly bound by the nucleus as the orbit is very closer to the nucleus.

In case of carbon, the fourth electron is present in the valence shell.The radius of valence shell is not so close as compared to lithium.The screening effect is also more for carbon as compared to lithium.

Hence, the energy required to remove a second electron from lithium is more that the energy required to remove fourth electron from carbon.



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Answer:

D) shrivel up, since the atmosphere exerts more force on the can as it cools.

Explanation:

As the water in the can is boiled the can gets heated up and contains hot vapour and gases which are rare in density and are in their expanded state. In this state when the can is sealed tightly such that no air leaks in or out of the can. When the temperature of the can drops, the gases shrink in volume and the pressure inside the can become less than the pressure of the atmosphere which leads to shriveling of the can.

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3 years ago
A beam of red light is made to pass through two slits that are 3.55 E-3 meters apart. On a screen 2.25 meters away from the slit
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I am assuming you know the relation obtained between slit width, distance of screen from slits, distance of interference pattern obtained on the screen from the center and the wavelength of monochromatic light used in Young's Double Slit experiment.
λ =\frac{y*d}{D} =  \frac{3.55*10^{-3}*1.25*10^{-4}  }{2.25} = 1.97*10^{-7} m
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7 0
2 years ago
In half wave rectifier circuit the diode and load resistance are connected in ...to ac power source​
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Working of a Half wave rectifier
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3 years ago
A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl
Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta is the angle of scattering.

Given that, the scattering angle is, \theta=147^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.

Here, the photon's incident wavelength is \lamda=2.78pm

Therefore,

\lambda^{'}=2.78+4.45=7.23 pm

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

where,\vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda,

and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therefore,

\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm

This is the de Broglie wavelength of the electron after scattering.

6 0
3 years ago
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