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beks73 [17]
3 years ago
14

The allowed energies of a simple atom are 0.0 eV, 4.0 eV, and 6.0 eV. Part A What wavelength(s) appear(s) in the atom's emission

spectrum
Physics
1 answer:
Dmitriy789 [7]3 years ago
3 0

Answer:

3.1 × 10^- 7 m and 2.1 × 10^-7 m

Explanation:

First we must convert each value of energy to Joules by multiplying its value by 1.6 ×10^-19. After that, we can now obtain the wavelength from E= hc/λ

Where;

h= planks constant

c= speed of light

λ= wavelength of light

For 6.0ev;

E= 6.0 × 1.6 ×10^-19

E= 9.6 × 10^-19 J

From

E= hc/λ

λ= hc/E

λ= 6.6 × 10^-34 × 3 × 10^8/9.6 × 10^-19

λ= 2.1 × 10^-7 m

For 4.0 eV

4.0 × 1.6 × 10^-19 = 6.4 × 10^-19 J

E= hc/λ

λ= hc/E

λ= 6.6 × 10^-34 × 3 × 10^8/6.4 × 10^-19

λ= 3.1 × 10^- 7 m

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Give two examples where friction is a nuisance
dlinn [17]

Answer:

it can cause wear and tear and reduces efficiency as energy is lost.

Explanation:

8 0
4 years ago
Un avión da "una vuelta mortal" de radio R = 500 m con una celeridad constante v = 360 km/h. Halla la fuerza ejercida por el asi
Airida [17]

Answer:

1400 N

Explanation:

Verá, durante el salto mortal, el piloto se mueve en una trayectoria circular y la fuerza que actúa sobre él es una fuerza centrípeta.

Sea la fuerza centrípeta F, la masa del piloto (m) = 70 Kg, el radio (r) = 500 my la velocidad (v) = 360 km / hr * 1000/3600 = 100 m / s

F = mv ^ 2 / r

F = 70 * (100) ^ 2/500

F = 1400 N

8 0
3 years ago
Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What p
Temka [501]

Answer:

The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

Explanation:

From General gas equation.

PV = nRT...............................  Equation 1

Where n = number of moles, V = volume, P = pressure, T = temperature, P = pressure, V = volume.

n = mass/molar mass .................. Equation 2

substituting equation 2 into equation 1.

PV = (mass/molar mass)RT

⇒ Mass/molar mass = PV/RT..................... Equation 3

But mass = Density × Volume

⇒ M = D × V.................... Equation 4

Where D = density, M = mass

Substituting equation 4 into equation 3

DV/molar mass = PV/RT............ Equation 5

Dividing both side of the equation by Volume (V) in Equation 5

D/molar mass = P/RT .............. Equation 6

Cross multiplying equation 6

D × RT = P × molar mass

∴ Molar mass = (D × RT)/P.................. Equation 7

Where D = 0.518 g/L , R = 0.0821 atm dm³/K.mol,

T = 25°C = 25 + 273 = 298 K,

P =721 mmHg = (721/760) atm= 0.949 atm

Substituting these values into equation 7

Molar mass = (0.518 × 0.0821 × 298)/0.949

Molar mass = 13.35 g/mole

The molar mass of the mixture is =13.35 g/mole

Let y be the mole fraction of Helium and 1-y be the mole fraction of oxygen.

∴ 13.35 = 4(y) + 32(1-y)

13.35 = 4y + 32 - 32y

Collecting like terms in the equation,

32y - 4y = 32 - 13.35

28y = 18.65

y = 18.65/28

y =0.666

y = 0.666 × 100 = 66.6%

∴The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

6 0
3 years ago
An oscillator consisting of a material point of m = 200g vibrates under the action of an elastic force according to the equation
Debora [2.8K]

Answer:

an oscillator consisting of a material point of m = 200g vibrates under the action of an elastic force according to the equation y = 0.2sin (π / 4 • t + π / 16) (m)

calculate the period and frequency of the oscillation

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8 0
3 years ago
According to the law of multiple proportions, if 12 g of carbon combine with 16 g of oxygen to form co, the number of grams of c
Ludmilka [50]

The number of grams of carbon that combine with 16 g of oxygen in the formation of CO₂ is 6g.

When two elements combine to make more than one compound, the masses of one element combined with a fixed amount of another element are in the ratio of whole numbers, according to the law of multiple proportions.

When combined with oxygen, carbon can produce two different compounds. They are referred to as carbon dioxide (CO₂) and carbon monoxide (CO).

Carbon monoxide is formed by combining 12 g of carbon with 16 g of oxygen whereas Carbon dioxide is formed when 12 g of carbon reacts with 32 g of oxygen. The amount of carbon is fixed at 12 g in each case. The mass ratio of carbon monoxide to carbon dioxide is 16: 32, or 1: 2.

But in the given case, 16g of oxygen is reacting instead of 32g. Therefore, the number of grams of carbon reacting will be:

\frac{12}{2}=6g

Thus, 6g of carbon will react with 16g of oxygen to form carbon dioxide.

Read more about Law of Multiple Proportions:

brainly.com/question/13058110

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7 0
1 year ago
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