Answer:
or 0.32 μm.
Explanation:
Given:
The radiations are UV radiation.
The frequency of the radiations absorbed (f) = 
The wavelength of the radiations absorbed (λ) = ?
We know that, the speed of ultraviolet radiations is same as speed of light.
So, speed of UV radiation (v) = 
Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:
Now, expressing the above equation in terms of wavelength 'λ', we have:
Now, plug in the given values and solve for 'λ'. This gives,
![\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B3%5Ctimes%2010%5E8%5C%20m%2Fs%7D%7B9.38%5Ctimes%2010%5E%7B14%7D%5C%20Hz%7D%5C%5C%5C%5C%5Clambda%3D3.2%5Ctimes%2010%5E%7B-7%7D%5C%20m%5C%5C%5C%5C%5Clambda%3D3.2%5Ctimes%2010%5E%7B-7%7D%5Ctimes%2010%5E%7B6%7D%5C%20%5Cmu%20m%5C%20%5B1%5C%20m%3D10%5E6%5C%20%5Cmu%20m%5D%5C%5C%5C%5C%5Clambda%3D3.2%5Ctimes%2010%5E%7B-1%7D%3D0.32%5C%20%5Cmu%20m)
Therefore, the wavelength of the radiations absorbed by the ozone is nearly or 0.32 μm.
Answer:
![F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]](https://tex.z-dn.net/?f=F_T%3D6k%5Cfrac%7BQ%5E2%7D%7BL%7D%5Chat%7Bi%7D%2B10k%5Cfrac%7BQ%5E2%7D%7BL%7D%5Chat%7Bj%7D%3D2k%5Cfrac%7BQ%5E2%7D%7BL%7D%5B3%5Chat%7Bi%7D%2B5%5Chat%7Bj%7D%5D)
Explanation:
I attached an image below with the scheme of the system:
The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:
![F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]](https://tex.z-dn.net/?f=F_T%3DF_Q%2BF_%7B3Q%7D%2BF_%7B4Q%7D%5C%5C%5C%5CF_T%3Dk%5Cfrac%7B%28Q%29%282Q%29%7D%7BR_1%7D%5Chat%7Bi%7D%2Bk%5Cfrac%7B%283Q%29%282Q%29%7D%7BR_2%7D%5Chat%7Bj%7D%2Bk%5Cfrac%7B%284Q%29%282Q%29%7D%7BR_3%7D%5Bcos%5Ctheta%20%5Chat%7Bi%7D%2Bsin%5Ctheta%20%5Chat%7Bj%7D%5D)
the distances R1, R2 and R3, for a square arrangement is:
R1 = L
R2 = L
R3 = (√2)L
θ = 45°
![F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]](https://tex.z-dn.net/?f=F_T%3Dk%5Cfrac%7B2Q%5E2%7D%7BL%7D%5Chat%7Bi%7D%2Bk%5Cfrac%7B6Q%5E2%7D%7BL%7D%5Chat%7Bj%7D%2Bk%5Cfrac%7B8Q%5E2%7D%7B%5Csqrt%7B2%7DL%7D%5Bcos%2845%5C%C2%B0%29%5Chat%7Bi%7D%2Bsin%2845%5C%C2%B0%29%5Chat%7Bj%7D%5D%5C%5C%5C%5CF_T%3Dk%5Cfrac%7B2Q%5E2%7D%7BL%7D%5Chat%7Bi%7D%2Bk%5Cfrac%7B6Q%5E2%7D%7BL%7D%5Chat%7Bj%7D%2Bk%5Cfrac%7B8Q%5E2%7D%7B%5Csqrt%7B2%7DL%7D%5B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Chat%7Bi%7D%2B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Chat%7Bj%7D%5D%5C%5C%5C%5CF_T%3D6k%5Cfrac%7BQ%5E2%7D%7BL%7D%5Chat%7Bi%7D%2B10k%5Cfrac%7BQ%5E2%7D%7BL%7D%5Chat%7Bj%7D%3D2k%5Cfrac%7BQ%5E2%7D%7BL%7D%5B3%5Chat%7Bi%7D%2B5%5Chat%7Bj%7D%5D)
and the magnitude is:
the direction is:
Answer:
b. negative
Explanation:
neutrons have a negative charge and protons have a proton has a positive charge
Answer:
a) 4.04*10^-12m
b) 0.0209nm
c) 0.253MeV
Explanation:
The formula for Compton's scattering is given by:
where h is the Planck's constant, m is the mass of the electron and c is the speed of light.
a) by replacing in the formula you obtain the Compton shift:
b) The change in photon energy is given by:
c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.
