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N76 [4]
3 years ago
7

The ground state energy of an oscillating electron is 1.23 eV. How much energy must be added to the electron to move it to the t

hird excited state? The fourth excited state?
Physics
1 answer:
Vikki [24]3 years ago
5 0

Answer:

  • The energy that must be added to the electron to move it to the third excited state is  -1.153 eV
  • The energy that must be added to the electron to move it to the fourth excited state is  -1.181 eV

Explanation:

Given;

Energy of electron in ground state (n = 1 ) = 1.23 eV

E₁ = 1.23 eV

Eₙ = E₁ /n²

where;

E₁ is the energy of the electron in ground state

n is the energy level,

For third excited state, n = 4

E₄ = E₁ /4²

E₄ = (1.23 eV) / 16

E₄ = 0.077 eV

Change in energy level, = E₄ - E₁ = 0.077 eV - 1.23 eV = -1.153 eV

The energy that must be added to the electron to move it to the third excited state is  -1.153 eV

For fourth excited state, n = 5

E₅ = E₁ /5²

E₄ = (1.23 eV) / 25

E₄ = 0.049 eV

Change in energy level, = E₅ - E₁ = 0.049 eV - 1.23 eV = -1.181 eV

The energy that must be added to the electron to move it to the fourth excited state is  -1.181 eV

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A body weighs 100newtons when submerged in water. calculate the upthrust action on the body​
Andrews [41]

Answer:

Upthrust = 20 N

Explanation:

The question says that "A body weighs 100N in air and 80N when submerged in water. Calculate the upthrust acting on the body ?"

Upthrust is defined as the force when a body is submerged in liquid, then liquid applies a force on it.

ATQ,

Weight of body in air is 100 N

Weight of body in water is 80 N

Upthrust is equal to the weight of body in air minus weight of body in water.

Upthrust = 100 N - 80 N

Upthrust = 20 N

So, 20 N of upthrust is acting on the body.

7 0
3 years ago
1. A 700 kg satellite is in orbit 2.4 x 106 meters from the center of the Earth. What is the force of
nevsk [136]

Answer:

4.8 \cdot 10^4 N

Explanation:

The force of gravity acting on the satellite is given by:

F=\frac{GMm}{r^2}

where

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

m is the mass of the satellite

r is the distance of the satellite from the Earth's centre

Here we have

m = 700 kg

r=2.4\cdot 10^6 m

Substituting into the equation, we find:

F=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(700)}{(2.4\cdot 10^6)^2}=4.8 \cdot 10^4 N

<em>Note that the distance mentioned in the problem (2.4 x 10^6 meters) is not realistic, since it is less than the radius of the Earth (6.37 x 10^6 meters).</em>

3 0
3 years ago
How do sinkholes occur? Overburden falls into the space left behind when bedrock is dissolved. Bedrock falls under the pressure
BabaBlast [244]

Answer: over burden is dissolved by water wind and acids

6 0
2 years ago
What does it mean for an element to be in a solid state?
MatroZZZ [7]
Solid elements are rigid elements. For example an element iron is in solid form. You can touch it and it’s hard.
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3 years ago
The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge
lesya [120]

Answer:

Loss, \Delta E=-10.63\ J

Explanation:

Given that,

Mass of particle 1, m_1=m =0.66\ kg

Mass of particle 2, m_2=7.4m =4.884\ kg

Speed of particle 1, v_1=2v_o=2\times 6=12\ m/s

Speed of particle 2, v_2=v_o=6\ m/s

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

m_1v_1+m_2v_2=(m_1+m_2)V

V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}

V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}

V = 6.71 m/s

Initial kinetic energy before the collision,

K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)

K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)

K_i=135.43\ J

Final kinetic energy after the collision,

K_f=\dfrac{1}{2}(m_1+m_2)V^2

K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2

K_f=124.80\ J

Lost in kinetic energy,

\Delta K=K_f-K_i

\Delta K=124.80-135.43

\Delta E=-10.63\ J

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0
3 years ago
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