Answer:
486nm
Explanation:
in order for an electron to transit from one level to another, the wavelength emitted is given by Rydberg Equation which states that
![\frac{1}{wavelength}=R.[\frac{1}{n_{f}^{2} } -\frac{1}{n_{i}^{2} }] \\n_{f}=2\\n_{i}=4\\R=Rydberg constant =1.097*10^{7}m^{-1}\\subtitiute \\\frac{1}{wavelength}=1.097*10^{7}[\frac{1}{2^{2} } -\frac{1}{4^{2}}]\\\frac{1}{wavelength}= 1.097*10^{7}*0.1875\\\frac{1}{wavelength}= 2.06*10^{6}\\wavelength=4.86*10{-7}m\\wavelength= 486nm\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bwavelength%7D%3DR.%5B%5Cfrac%7B1%7D%7Bn_%7Bf%7D%5E%7B2%7D%20%7D%20-%5Cfrac%7B1%7D%7Bn_%7Bi%7D%5E%7B2%7D%20%7D%5D%20%5C%5Cn_%7Bf%7D%3D2%5C%5Cn_%7Bi%7D%3D4%5C%5CR%3DRydberg%20constant%20%3D1.097%2A10%5E%7B7%7Dm%5E%7B-1%7D%5C%5Csubtitiute%20%5C%5C%5Cfrac%7B1%7D%7Bwavelength%7D%3D1.097%2A10%5E%7B7%7D%5B%5Cfrac%7B1%7D%7B2%5E%7B2%7D%20%7D%20-%5Cfrac%7B1%7D%7B4%5E%7B2%7D%7D%5D%5C%5C%5Cfrac%7B1%7D%7Bwavelength%7D%3D%201.097%2A10%5E%7B7%7D%2A0.1875%5C%5C%5Cfrac%7B1%7D%7Bwavelength%7D%3D%202.06%2A10%5E%7B6%7D%5C%5Cwavelength%3D4.86%2A10%7B-7%7Dm%5C%5Cwavelength%3D%20486nm%5C%5C)
Hence the photon must possess a wavelength of 486nm in order to send the electron to the n=4 state
Daniddmelo says it right there, don't know why he got reported.
The potential energy (PE) is mass x height x gravity. So it would be 25 kg x 4 m x 9.8 = 980 joules. The child starts out with 980 joules of potential energy. The kinetic energy (KE) is (1/2) x mass x velocity squared. KE = (1/2) x 25 kg x 5 m/s2 = 312.5 joules. So he ends with 312.5 joules of kinetic energy. The Energy lost to friction = PE - KE. 980- 312.5 = 667.5 joules of energy lost to friction.
Please don't just copy and paste, and thank you Dan cause you practically did it I just... elaborated more? I dunno.
Answer:
X₃₁ = 0.58 m and X₃₂ = -1.38 m
Explanation:
For this exercise we use Newton's second law where the force is the Coulomb force
F₁₃ - F₂₃ = 0
F₁₃ = F₂₃
Since all charges are of the same sign, forces are repulsive
F₁₃ = k q₁ q₃ / r₁₃²
F₂₃ = k q₂ q₃ / r₂₃²
Let's find the distances
r₁₃ = x₃- 0
r₂₃ = 2 –x₃
We substitute
k q q / x₃² = k 4q q / (2-x₃)²
q² (2 - x₃)² = 4 q² x₃²
4- 4x₃ + x₃² = 4 x₃²
5x₃² + 4 x₃ - 4 = 0
We solve the quadratic equation
x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2 5
x₃ = [-4 ± 9.80] 10
X₃₁ = 0.58 m
X₃₂ = -1.38 m
For this two distance it is given that the two forces are equal
Explanation:
The reading on the scale is
W = m(g + a)
= (77 kg)(9.8 m/s^2 + 2 m/s^2)
= 908.6 N
