To solve this question, we use the wave equation which is:
C=f*λ
where:
C is the speed;
f is the frequency;
λ is the wavelength
So in this case, plugging in our values in the problem. This will give us:
C = 261.6Hz × 1.31m
= 342.696 m/s is the answer.
I think D. It starts at (0.0) and goes to the correct points so it makes sense
Explanation:
Given that,
Wavelength = 6.0 nm
de Broglie wavelength = 6.0 nm
(a). We need to calculate the energy of photon
Using formula of energy
![E = \dfrac{hc}{\lambda}](https://tex.z-dn.net/?f=E%20%3D%20%5Cdfrac%7Bhc%7D%7B%5Clambda%7D)
![E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B6.63%5Ctimes10%5E%7B-34%7D%5Ctimes3%5Ctimes10%5E%7B8%7D%7D%7B6.0%5Ctimes10%5E%7B-9%7D%7D)
![E=3.315\times10^{-17}\ J](https://tex.z-dn.net/?f=E%3D3.315%5Ctimes10%5E%7B-17%7D%5C%20J)
(b). We need to calculate the kinetic energy of an electron
Using formula of kinetic energy
![\lambda=\dfrac{h}{\sqrt{2mE}}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bh%7D%7B%5Csqrt%7B2mE%7D%7D)
![E=\dfrac{h^2}{2m\lambda^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bh%5E2%7D%7B2m%5Clambda%5E2%7D)
Put the value into the formula
![E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B%286.63%5Ctimes10%5E%7B-34%7D%29%5E2%7D%7B2%5Ctimes9.1%5Ctimes10%5E%7B-31%7D%5Ctimes%286.0%5Ctimes10%5E%7B-9%7D%29%5E2%7D)
![E=6.709\times10^{-21}\ J](https://tex.z-dn.net/?f=E%3D6.709%5Ctimes10%5E%7B-21%7D%5C%20J)
(c). We need to calculate the energy of photon
Using formula of energy
![E = \dfrac{hc}{\lambda}](https://tex.z-dn.net/?f=E%20%3D%20%5Cdfrac%7Bhc%7D%7B%5Clambda%7D)
![E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B6.63%5Ctimes10%5E%7B-34%7D%5Ctimes3%5Ctimes10%5E%7B8%7D%7D%7B6.0%5Ctimes10%5E%7B-15%7D%7D)
![E=3.315\times10^{-11}\ J](https://tex.z-dn.net/?f=E%3D3.315%5Ctimes10%5E%7B-11%7D%5C%20J)
(d). We need to calculate the kinetic energy of an electron
Using formula of kinetic energy
![\lambda=\dfrac{h}{\sqrt{2mE}}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bh%7D%7B%5Csqrt%7B2mE%7D%7D)
![E=\dfrac{h^2}{2m\lambda^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bh%5E2%7D%7B2m%5Clambda%5E2%7D)
Put the value into the formula
![E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B%286.63%5Ctimes10%5E%7B-34%7D%29%5E2%7D%7B2%5Ctimes9.1%5Ctimes10%5E%7B-31%7D%5Ctimes%286.0%5Ctimes10%5E%7B-15%7D%29%5E2%7D)
![E=6.709\times10^{-9}\ J](https://tex.z-dn.net/?f=E%3D6.709%5Ctimes10%5E%7B-9%7D%5C%20J)
Hence, This is the required solution.
Speed of particle B is 2v₀/3 m/s to the left. Particle A and particle B will always have equal speed since they experience equal forces.
<h3>Conservation of energy</h3>
The speed and direction of the particle B is determined by applying the principle of conservation of energy as follows;
K.E₁ + P.E₁ = K.E₂ + P.E₂
![\frac{1}{2} Mv^2_A + \frac{G}{r^2} = \frac{1}{2} Mv^2_B + \frac{G}{r^2} \\\\ \frac{1}{2} Mv^2_A = \frac{1}{2} Mv^2_B\\\\v^2_A = v^2_B\\\\v_A = v_B](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20Mv%5E2_A%20%2B%20%5Cfrac%7BG%7D%7Br%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20Mv%5E2_B%20%2B%20%5Cfrac%7BG%7D%7Br%5E2%7D%20%5C%5C%5C%5C%20%5Cfrac%7B1%7D%7B2%7D%20Mv%5E2_A%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20Mv%5E2_B%5C%5C%5C%5Cv%5E2_A%20%3D%20v%5E2_B%5C%5C%5C%5Cv_A%20%3D%20v_B)
![v_B = \frac{2v_0}{3} \ m/s \ to \ the \ left](https://tex.z-dn.net/?f=v_B%20%3D%20%5Cfrac%7B2v_0%7D%7B3%7D%20%20%5C%20m%2Fs%20%5C%20to%20%5C%20the%20%5C%20left)
At any given position, the speed of particle A and particle B will be equal, since they experience equal force and they have equal masses.
The complete question is below:
Particle A and particle B, each of mass M, move along the x-axis exerting a force on each other. The potential energy of the system of two particles assosicated with the force is given by the equation U=G/r 2, where r is the distance between the two particles and G is a positive constant. At time t=T1 particle A is observed to be traveling with speed 2vo/3 to the left. The speed and direction of motion of particle B is ?
Learn more about conservation of energy here: brainly.com/question/166559
No. A mirror works because of reflection.