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Stels [109]
2 years ago
15

The ionization energy of singly charged anions _____ as electron affinity of the neutral atom _____

Physics
1 answer:
eimsori [14]2 years ago
3 0

Answer:

  • The ionization energy of an atom is the required amount of energy which is provided to the atom, so, that the required number of electrons are removed from outermost shells of the electronic configuration of the atom. It is minimum amount of energy that is required by the atom, so, that it can remove the charged particle from its outer shell.
  • The ionization energy of singly charged anions <u>less same</u> as electron affinity of the neutral atom for <u>the electrons.</u>
  • As, for the anion having the net charge on its body will have larger size then any other normal atom, as the atom will have large and it will be difficult for the nucleus to hold off its whole electronic configuration into one place. This process makes it more easy for the foreign source or agent to remove the required number of charged particles from the atom's outer region.
  • While it may be same as the phenomenon which occurs inside the neutral atom's whole structure, as there is less affinity for the electrons  and that is how the two of them are termed as inter related to one another in the mechanism of having the less affinity for the electrons.

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A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
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Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

Idt=0.00000157=1.57\times10^{-6}\ C

We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

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In an experiment to illustrate the propagation of sound through fluids, identical sound sources emitting at 440 Hz are used to p
earnstyle [38]

Answer:

B. In mercury, the frequency of the wave is the same as in ethanol, but the wavelength is greater.

Explanation:

To solve this easily, we can just calculate the wavelength of the sound in Ethanol and in Mercury.

In Ethanol, the wavelength will be:

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