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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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Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.
olga2289 [7]

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NO_{1.499}

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8 0
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Read 2 more answers
Masses A and B rest on very light pistons that enclose a fluid.There is no friction between the pistons and the cylinders they f
RSB [31]

Answer:

D)Not enough information

Explanation:

According to Pascal's principle, the pressure exerted on the two pistons is equal:

p_A = p_B

Pressure is given by the ratio between force F and area A, so we can write

\frac{F_A}{A_A}=\frac{F_B}{A_B}

The force exerted on each piston is just equal to the weight of the corresponding mass: F=W=mg, where m is the mass and g is the gravitational acceleration. So the equation becomes

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Now we can rewrite the mass as the product of volume, V, times density, d:

\frac{V_A d_A g}{A_A}=\frac{V_B d_B g}{A_B}

We also know that A_B = 2.0 m^2\\A_A = 1.0 m^2

So we can further re-arrange the equation (and simplify g as well):

\frac{V_A d_A}{1}=\frac{V_B d_B}{2}

\frac{d_A}{d_B}=\frac{V_B}{2V_A}

We are also told that block B has bigger volume than block A: V_B > V_A. However, this information is not enough to allow us to say if the fraction on the right is greater than 1 or smaller than 1: therefore, we cannot conclude anything about the densities of the two objects.

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