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Natasha2012 [34]
3 years ago
10

According to the equation for the Balmer line spectrum of hydrogen, a value of n = 3 gives a red spectral line at 656.3 nm, a va

lue of n = 4 gives a green line at 486.1 nm, and a value of n = 5 gives a blue line at 434.0 nm. Calculate the energy in kilojoules per mole of the radiation corresponding to each of these spectral lines. McMurry, John E.. Chemistry (p. 189). Pearson Education. Kindle Edition.
Physics
1 answer:
nirvana33 [79]3 years ago
7 0

Answer:

E3 = 3.03 10⁻¹⁶ kJ,  E4 = 4.09 10⁻¹⁶ kJ and  E5 = 4.58 10⁻¹⁶ kJ

Explanation:

They give us some spectral lines of the Balmer series, let's take the opportunity to place the values in SI units

     n = 3        λ = 656.3 nm = 656.3 10⁻⁹ m

     n = 4        λ = 486.1 nm = 486.1 10⁻⁹ m

     n = 5        λ=434.0 nm = 434.0 10⁻⁹ m

Let's use the Planck equation

     E = h f

The speed of light equation

   c = λ f

replace

    E = h c /λ

Where h is the Planck constant that is worth 6.63 10⁻³⁴ J s and c is the speed of light that is worth 3 10⁸ m / s

Let's calculate the energies

     E = 6.63 10⁻³⁴ 3 10⁸ / λ

     E = 19.89 10⁻²⁶ /λ

n = 3

    E3 = 19.89 10⁻²⁶ / 656.3 10⁻⁹

    E3 = 3.03 10⁻¹⁹ J

    1 kJ = 10³ J

    E3 = 3.03 10⁻¹⁶ kJ

n = 4

    E4 = 19.89 10⁻²⁶ /486.1 10⁻⁹

    E4 = 4.09 10⁻¹⁹ J

    E4 = 4.09 10⁻¹⁶ kJ

n = 5

    E5 = 19.89 10⁻²⁶ /434.0 10⁻⁹

    E5 = 4.58 10⁻¹⁹ J

    E5 = 4.58 10⁻¹⁶ kJ

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Ok so use trigonometry to work out the vertical component of velocity.

sin(25) =opp/hyp
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30*sin(25) which equals 12.67ms^-1

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plug in some numbers and do some maths and we get 2.583s

this is the total air time of the golf ball.

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2 years ago
Which of the following phenomena support the idea that light is a wave?
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The magnitude of the electric current is directly proportional to the _____ of the electric field.
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Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax
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1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its x-component is

F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its x-component is

F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N

So, the x-component of the resultant force is

F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its y-component is

F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

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Therefore its y-component is

F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N

So, the y-component of the resultant force is

F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N

3)

The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.

Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:

F=\sqrt{F_x^2+F_y^2}

Where in this problem, we have:

F_x=-7.14 N is the x-component

F_y=2.70 N is the y-component

And substituting, we find:

F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N

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3 years ago
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Answer:

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Explanation:

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Using the formula again

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