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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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By wave particle  duality.

Wavelength , λ = h / mv

where h = Planck's constant = 6.63 * 10⁻³⁴ Js,  m = mass in kg,  v = velocity in m/s.
m = 1kg,  v = 4.5 m/s

λ = h / mv

λ = (6.63 * 10⁻³⁴) /(1*4.5)

λ ≈  1.473 * 10⁻³⁴  m

Option D.
7 0
3 years ago
A solid spherical conductor has a radius of 12 cm. The electric field at 24 from the center of the sphere has a magnitude of 640
hoa [83]

Answer:

Charge density on the sphere = 2.2 × 10⁻⁸ C/m²

Explanation:

Given:

Radius of sphere (r) = 12 cm = 0.12 m

Distance from the electric field R = 24 cm = 0.24 m

Magnitude (E) = 640 N/C

Find:

Charge density on the sphere

Computation:

Charge on the sphere (q) = (1/K)ER²            (K = 9 × 10⁹)

Charge on the sphere (q) = [1/(9 × 10⁹)](640)(0.24)²

Charge on the sphere (q) = 4 × 10⁻⁹ C

Charge density on the sphere = q / [4πr²]

Charge density on the sphere = [4 × 10⁻⁹] / [4(3.14)(0.12)²]

Charge density on the sphere = [4 × 10⁻⁹] / [0.18]

Charge density on the sphere = 2.2 × 10⁻⁸ C/m²

4 0
3 years ago
Which provides the best analogy for an electron in an atomic orbital?
Svetllana [295]
<h2>Answer: a bee trying to escape from a closed jar  </h2>

In an atom the electrons will occupy orbitals so that their energy is as small as possible. That is why the orbitals are ordered based on their energy level in an increasing order, which is associated with a particular range of energy based on its distance from the atom nucleus.

In this sense, an electron "jumps" from one level to another in the atom in the same way a bee tries to escape from a closed jar.

3 0
3 years ago
a mass on a spring vibrates in simple harmonic motion at an amplitude of 8.0 cm. if the mass of the object is 0.20kg and the spr
Reil [10]

Answer:

4.06 Hz

Explanation:

For simple harmonic motion, frequency is given by

f=\frac {1}{2\pi}\times \sqrt{\frac {k}{m}} where k is spring constant and m is the mass of the object.

Substituting 0.2 Kg for mass and 130 N/m for k then

f=\frac {1}{2\pi}\times \sqrt{\frac {130}{0.2}}=4.057670803\\f\approx 4.06 Hz

5 0
3 years ago
What do these letters stand for<br> P=mv
victus00 [196]

Answer:

The equation for momentum of a piece of matter.

In either case, the momentum would be less than a linebacker hitting you at full speed. The equation for momentum is written: p = mv where p stands for momentum. That is, mass times velocity equals momentum.

Explanation:

Hope This Helps

Have A Great Day

4 0
2 years ago
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