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harkovskaia [24]
3 years ago
15

The first excited state of a particular atom in a gas is 5.7 eV above the ground state. A moving electron collides with one of t

hese atoms, and excites the atom to its first excited state. Immediately after the collision the kinetic energy of the electron is 3.4 eV. What was the kinetic energy of the electron just before the collision?
Physics
1 answer:
Georgia [21]3 years ago
5 0

Answer:

Kinetic energy, E = 9.1 eV

Explanation:

It is given that,

At the first excited state of a particular atom is 5.7 eV above the ground state. The kinetic energy of the atom after the collision is 3.4 eV.

We need to find the kinetic energy of the electron just before the collision. The conservation of momentum will be followed here. The energy gets transferred from moving electrons to these atoms is 5.7.

Let E is the kinetic energy of the electron just before the collision. It is equal to :

E = 5.7 eV + 3.4 eV

E = 9.1 eV

So, the kinetic energy of the electron just before the collision is 9.1 eV. Hence, this is the required solution.

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2 years ago
A series L-R-C circuit consists of a 226 Ω resistor, a 27.4 mH inductor, a 11.55 µF capacitor, and an AC source of amplitude 15
DanielleElmas [232]

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Explanation:

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6 0
3 years ago
Where does denitrification happen?
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The correct anwser is in deep soil or anoxic aquatic sediment.
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A grinding wheel is a uniform cylinder with a radius of 8.65 cm and a mass of 0.400 kg . Calculate its moment of inertia about i
anyanavicka [17]

Answer:

I = 1.5*10⁻³ kg*m²

Explanation:

  • It can be showed that the moment of inertia (or rotational inertia) for a uniform cylinder of mass m and radius r, respect an longitudinal axis going through its center (parallel to the height of the cylinder) can be written as follows:

       I = \frac{1}{2}*m*r^{2}  = \frac{1}{2}*0.400 kg*(0.0865m)^{2}  = 1.5e-3 kg*m2

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3 years ago
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