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madreJ [45]
3 years ago
7

Collisions between atoms are often elastic, but sometimes inelastic collisions occur, and the lost kinetic energy can become int

ernal energy in the atoms, exciting electrons to higher energy orbitals or even ejecting then from the atom entirely, which is called ionization (and the required energy the ionization energy).2 (a) A Cs 133 atom (mass 133 amu) has an ionization energy of 4.0 eV.3 Express this energy in joules. (b) What is the minimum speed that a O16 atom can have if it is to ionize a cesium atom at rest upon impact?
Physics
1 answer:
Galina-37 [17]3 years ago
4 0

Answer:

a) E = 6.4 1019 J    b)  v = 0.69 10⁴4 m / s

Explication

a) convert E = 4.0 eV

    1 eV = 1.6 10⁻¹⁹ J

   E = 4.0 eV (1.6 10⁻¹⁹ J / 1 eV)

   E = 6.4 10⁻¹⁹ J

b) Suppose we have a frontal shock and all the kinetic energy of oxygen is transferred to Cs

    Ei = K = ½ m v²

    Ef = 6.4 10⁻¹⁹ J

    ½ m v² = 6.4 10⁻¹⁹

The oxygen mass of the periodic table is

     PA = 15,999 u

     1u = 1.660 10⁻²⁷ kg

     Pa = 15,999 1,660 10⁻²⁷ kg

     m= Pa = 26,558 10⁻²⁷ kg

Let's calculate the speed

    v2 = 2 / m 6.4 10⁻¹⁹

    v2 = 2 / 26,558 10⁻²⁷ 6.4 10⁻¹⁹ =

    v = √0.4819 10⁸

    v = 0.69 10⁴4 m / s

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Answer:

1.232\times 10^{-5}\ T.

Explanation:

<u>Given:</u>

  • Current through the wire, passing through the origin, I_1 = 250\ A.
  • Current through the wire, passing through the y axis, r_y=1.8\ m., I_2 = 50\ A.

According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to \mu_o times the net current threading the loop.

\oint \vec B \cdot d\vec l=\mu_o I.

In case of a circular loop, the directions of magnetic field and the line element d\vec l, both are along the tangent of the loop at that point, therefore, \vec B\cdot d\vec l = B\ dl.

\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\

\oint dl is the circumference of the Amperian loop = 2\pi r

Therefore,

B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.

It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, r_1 = 3.510\ m.

The magnetic field at the given point due to this wire is given by:

B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.

For the first wire, passing through the y-axis:

Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m,  concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, r_2 = 5.310\ m.

The magnetic field at the given point due to this wire is given by:

B_2 = \dfrac{\mu_o I_2}{2\pi r_2}\\=\dfrac{4\pi \times 10^{-7}\times 50}{2\pi \times 5.310}=1.88\times 10^{-6}\ T.

The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.

Thus, the net magnetic field at r_r=-3.510\ m is given by

B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.

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