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Ann [662]
3 years ago
11

Heated lithium atoms emit photons of light with an energy of 2.961 × 10−19 J. Calculate the frequency and wavelength of one of t

hese photons. What is the total energy in 1 mole of these photons? What is the color of the emitted light?
Physics
1 answer:
tatiyna3 years ago
6 0

Answer:

4.5 x 10¹⁴ Hz

666.7 nm

1.8 x 10⁵ J

The color of the emitted light is red

Explanation:

E = energy of photons of light = 2.961 x 10⁻¹⁹ J

f = frequency of the photon

Energy of photons is given as

E = h f

2.961 x 10⁻¹⁹ = (6.63 x 10⁻³⁴) f

f = 4.5 x 10¹⁴ Hz

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of photon

Using the equation

c = f λ

3 x 10⁸ = (4.5 x 10¹⁴) λ

λ = 0.6667 x 10⁻⁶ m

λ = 666.7 x 10⁻⁹ m

λ = 666.7 nm

n = number of photons in 1 mole = 6.023 x 10²³

U = energy of 1 mole of photons

Energy of 1 mole of photons is given as

U = n E

U = (6.023 x 10²³) (2.961 x 10⁻¹⁹)

U = 1.8 x 10⁵ J

The color of the emitted light is red

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NASA is designing a Mars-lander that will enter the Martian atmosphere at high speed. To land safely it must slow to a constant
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Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

c) area of the parachute should be 12.68 m²

Explanation:

Given the data in the question;

V = 20 m/s

A = 200 m²

drag co-efficient CD = 1.855

g = 3.71 m/s²

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a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²

FD = 742 N

we know that;

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Fg = gravity force

Fg = mg

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FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;

Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

FD = 1780.8 N

Now, FD = 1/2 × CD × β × A × V², we solve for A

A = FD / 0.5 × CD × β × V²

we substitute

A = 1780.8  / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

A = 480 m²

Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

g = 9.8 m/s²,

β" = 1 kg/m³

v" = 20 m/s

M" = 480 kg

we know that;

FD = Fg = M"g

FD = 480 kg × 9.8 m/s² = 4704 N

from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

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