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enot [183]
3 years ago
5

When the electron in a hydrogen atom moves from n = 6 to n = 1, light with a wavelength of ________ nm is emitted?

Physics
1 answer:
Oduvanchick [21]3 years ago
3 0
First we find the energy level with the following formula, where a is the energy level, n1 is the final energy level, n2 is the starting energy level and r is Rydberg's constant in Joules
a = r \times ( \frac{1}{n1}  -  \frac{1}{n2} )
We insert the values

a = 2.18 \times  {10}^{ - 18}  \times ( \frac{1}{ {1}^{2} } -  \frac{1}{ {6}^{2} }  )
= 2.12 \times {10}^{ - 18}
The wavelength is found with this formula, where h is Planck's constant and c is the speed of light
wavelength =  \frac{h \times c}{a}
Finally we insert the values
\frac{6.626 \times  {10}^{ - 34}  \times 3 \times  {10}^{8} }{2.12 \times  {10}^{ - 18} }  = 9.376 \times  {10}^{ - 8}
Which is the same as 93.8 nm
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Water flows past a flat plate that is oriented parallel to the flow with an upstream velocity of 0.4 m/s. (a) Determine the appr
Trava [24]

Answer:

1.12 m

0.08291 m

Explanation:

u = Upstream velocity = 0.4 m/s

Re = Reynold's number = 5\times 10^5 (turbulent)

\nu = Viscosity of water = 1.12\times 10^{-6}\ Pas

Here the flow is turbulent so we have the relation

Re_{xcr}=\frac{ux_{cr}}{\nu}\\\Rightarrow x_{cr}=\frac{Re_{xcr}\nu}{u}\\\Rightarrow x_{cr}=\frac{5\times 10^5\times 1.12\times 10^{-6}}{0.4}\\\Rightarrow x_{cr}=1.4\ m

The approximate location downstream from the leading edge where the boundary layer becomes turbulent is 1.4 m

Boundary layer thickness relation is given by

\delta={\frac{\nu x}{u}}^{\frac{1}{5}}\\\Rightarrow \delta={\frac{1.12\times 10^{-6}\times 1.4}{0.4}}^{\frac{1}{5}}\\\Rightarrow \delta=0.08291\ m

The boundary layer thickness is 0.08291 m

4 0
4 years ago
A velocity-time graph is shown below: (2 points)
BabaBlast [244]

Answer:

2m/s²

Explanation:

  • V=Vstart+at
  • rewrite that to find a so a=(V-Vstart)/t

part A of graph

  • a=(20m/s–0m/s)/5s
  • a=4m/s²

part B of graph

  • a=(0m/s–0m/s)/5s
  • a=0m/s²

the average between the two is both answers added divided by the number of answers

  • (4m/s²–0m/s²)/2
  • 4m/s²/2
  • 2m/s²
5 0
3 years ago
I NEED HELP ON THIS QUESTION!
GarryVolchara [31]

Forward because things that are in motion stay in motion.

Hope that helped!

~Izzy

4 0
3 years ago
Read 2 more answers
A hot air balloon is filled with 1.45 × 10 6 L of an ideal gas on a cool morning ( 11 ∘ C ) . The air is heated to 109 ∘ C . Wha
just olya [345]

Answer:

<em>The volume of air after the balloon is heated = 1.95 × 10⁶ L</em>

Explanation:

Charles Law: Charles' law states that the volume of a given mass of gas is directly proportional to the temperature in Kelvin, provided that the pressure remains constant.

It can be expressed mathematically as,

V₁/T₁ = V₂/T₂

Making V₂ The subject of the equation

V₂ = (V₁/T₁)T₂.................... Equation 1

Where V₁ = Initial Volume, T₁ = Initial Temperature, V₂ = Final Volume, T₂ = final Temperature

<em>Given: V₁ = 1.45 × 10⁶ L, T₁ = 11 °C = (11 + 273) K = 284 K, T₂ = 109 °C = (109 + 273) = 382 K.</em>

<em>Substituting these values into equation 1 above,</em>

<em>V₂ = (1.45×10⁶)382/284</em>

<em>V₂ = 1.95 × 10⁶ L</em>

<em>Therefore the volume of air after the balloon is heated = 1.95 × 10⁶ L</em>

6 0
3 years ago
a particle that has a mass of 2.5 kg is moving in the positive X-direction with a constant velocity of 1.6m/s. Suddenly a consta
malfutka [58]
The particle has a constant horizontal velocity, and a vertical force won't affect the horizontal speed, so it should be fairly easy to find the last part, "the time taken for a 10m horizontal displacement," using a kinematic equation.
X = x + vt + (1/2)at²
10 = 0 + (1.6)t + (1/2)(0)t²
10/1.6 = t
t = 6.25s

So now we have to find the vertical displacement over 6.25 seconds on a particle of a 2.5kg mass with a force of 8N.
Start with Newton's second law.
F = ma
8 = (2.5)a
a = 3.2m/s²

Now, use kinematics again.
Y = y + vt + (1/2)at²
Y = 0 + (0)(6.25) + (1/2)(3.2)(6.25)²
Y = <u>62.5m</u>
7 0
3 years ago
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