Answer:
A λ = 97.23 nm
, B) λ = 486.2 nm
, C) λ = 53326 nm
Explanation:
With that problem let's use the Bohr model equation for the hydrogen atom
= -k e² /2a₀ 1/n²
For a transition between two states we have
- = -k e² /2a₀ (1/ ² - 1 / n₀²)
Now this energy is given by the Planck equation
E = h f
And the speed of light is
c = λ f
Let's replace
h c / λ = - k e² /2a₀ (1 / ² - 1 / no₀²)
1 / λ = - k e² /2a₀ hc (1 / ² -1 / n₀²)
Where the constants are the Rydberg constant = 1.097 10⁷ m⁻¹
1 / λ = (1 / n₀² - 1 / nf²)
Now we can substitute the given values
Part A
Initial state n₀ = 1 to the final state = 4
1 / λ = 1.097 10⁷ (1/1 - 1/4²)
1 / λ = 1.0284 10⁷ m⁻¹
λ = 9.723 10⁻⁸ m
We reduce to nm
λ = 9.723 10⁻⁸ m (10⁹ nm / 1m)
λ = 97.23 nm
Part B
Initial state n₀ = 2 final state = 4
1 / λ = 1.097 10⁷ (1/2² - 1/4²)
1 / λ = 0.2056 10⁻⁷ m
λ = 486.2 nm
Part C
Initial state n₀ = 3
1 / λ = 1,097 10⁷ (1/3² - 1/4²)
1 / λ = 5.3326 10⁵ m⁻¹
λ = 5.3326 10-5 m
λ = 53326 nm
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Answer:
very hard others will answer it
Explanation:
hard
Well they live in dry areas bc there jumps hole water
OK, the wedge is accelerating (a) at Theta = 180 degrees (to the right) and the wedge is inclined theta = 75 degrees. For the m = 2 kg block to remain at rest all we need is a net force f = W cos(theta) - F sin(theta) = 0; where F = ma and W = mg the weight of the block. That is, the weight component along the incline is offset by the acceleration component along the surface; so the block does not slide.
Solving we have W cos(theta) = mg cos(theta) = ma sin(theta) = F sin(theta); such that a = g cos(theta)/sin(theta) = g cot(theta). Assuming g ~ 9.81 m/sec^2, you can now plug and chug to find the answer.
<span>The physics is this...when the net force on a body is f = 0, that body will not accelerate and start to move if it is already still. So when the block's weight component along the surface of the wedge is offset by the equal but opposite force along the surface of the accelerating wedge, the still block will not move.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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