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pav-90 [236]
2 years ago
14

What quantities can be calculated from the bohr equation for the energy levels of the hydrogen atom?

Physics
1 answer:
patriot [66]2 years ago
6 0

We can find the energy needed to ionize the hydrogen atom.

We can find the wavelength of a spectral line.

We can find the energy change of the electron moving between two levels.

The first atomic model to adequately explain the radiation spectra of atomic hydrogen was Bohr's model of the hydrogen atom. The atomic Hydrogen model was first presented by Niels Bohr in 1913. Rutherford's model of the hydrogen atom leaves several holes, which Bohr's model of the hydrogen atom tries to fill in.

It has a particular place in history since it introduced the quantum theory, which led to the development of quantum mechanics. Bohr proposed that electrons moved in predetermined orbits or shells with defined radii around the nucleus. It was impossible for electrons to reside between any shells other than those having a radius given by the equation below.

r (n) =n^{2} * r(1)

E (n) = \frac{1}{n^{2} } 13.6eV

Learn more about hydrogen atom here;

brainly.com/question/8806577

#SPJ4

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agasfer [191]

Answer:

746 watts

Explanation:

7 0
2 years ago
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h
Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

v_{1}=\dfrac{3.0\times10^{3}}{3600}

v_{1}=0.833\ m/s

The force F₁is constant acceleration is also a constant.

F_{1}=ma_{1}

We need to calculate the acceleration

Using formula of acceleration

a_{1}=\dfrac{v}{t}

a_{1}=\dfrac{0.833}{10}

a_{1}=0.083\ m/s^2

Similarly,

F_{2}=ma_{2}

For total force,

F_{3}=F_{2}+F_{1}

ma_{3}=ma_{2}+ma_{1}

The speed of second tugboat is

v=\dfrac{11\times10^{3}}{3600}

v=3.05\ m/s

We need to calculate total acceleration

a_{3}=\dfrac{v}{t}

a_{3}=\dfrac{3.05}{10}

a_{3}=0.305\ m/s^2

We need to calculate the acceleration a₂

0.305=a_{2}+0.083

a_{2}=0.305-0.083

a_{2}=0.222\ m/s^2

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}

\dfrac{F_{1}}{F_{2}}=3.7

F_{1}=3.7F_{2}

Hence, The magnitude of F₁ is 3.7 times of F₂

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3 years ago
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A paper is placed in between two books can be quickly pulled out without movinf the books. whuch statement explains this phenome
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