Answer:
![r=5.278\times 10^{-4}\ m](https://tex.z-dn.net/?f=r%3D5.278%5Ctimes%2010%5E%7B-4%7D%5C%20m)
Explanation:
Given that:
- magnetic field intensity,
![B=0.07\ T](https://tex.z-dn.net/?f=B%3D0.07%5C%20T)
- kinetic energy of electron,
![KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J](https://tex.z-dn.net/?f=KE%3D1.2%5C%20eV%3D%201.2%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%5C%20J%3D%201.92%5Ctimes%2010%5E%7B-19%7D%5C%20J)
- we have mass of electron,
![m=9.1\times 10^{-31}\ kg](https://tex.z-dn.net/?f=m%3D9.1%5Ctimes%2010%5E%7B-31%7D%5C%20kg)
<em>Now, form the mathematical expression of Kinetic Energy:</em>
![KE= \frac{1}{2} m.v^2](https://tex.z-dn.net/?f=KE%3D%20%5Cfrac%7B1%7D%7B2%7D%20m.v%5E2)
![1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2](https://tex.z-dn.net/?f=1.92%5Ctimes%2010%5E%7B-19%7D%3D0.5%5Ctimes%209.1%5Ctimes%2010%5E%7B-31%7D%5Ctimes%20v%5E2)
![v^2=4.2198\times 10^{11}](https://tex.z-dn.net/?f=v%5E2%3D4.2198%5Ctimes%2010%5E%7B11%7D)
![v=6.496\times 10^6\ m.s^{-1}](https://tex.z-dn.net/?f=v%3D6.496%5Ctimes%2010%5E6%5C%20m.s%5E%7B-1%7D)
<u>from the relation of magnetic and centripetal forces we have the radius as:</u>
![r=\frac{m.v}{q.B}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bm.v%7D%7Bq.B%7D)
![r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B9.1%5Ctimes%2010%5E%7B-31%7D%5Ctimes%206.496%5Ctimes%2010%5E6%20%7D%7B1.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%200.07%7D)
![r=5.278\times 10^{-4}\ m](https://tex.z-dn.net/?f=r%3D5.278%5Ctimes%2010%5E%7B-4%7D%5C%20m)
Answer:
![\ m/s](https://tex.z-dn.net/?f=%3C3068.2352%2C%20800%2C%200%3E%5C%20m%2Fs)
Explanation:
F = Force = ![](https://tex.z-dn.net/?f=%3C-1.12%5Ctimes%2010%5E%7B-11%7D%2C%200%2C%200%3E)
m = Mass of proton = ![1.7\times 10^{-27\ kg](https://tex.z-dn.net/?f=1.7%5Ctimes%2010%5E%7B-27%5C%20kg)
t = Time taken = ![2\times 10^{-14}\ s](https://tex.z-dn.net/?f=2%5Ctimes%2010%5E%7B-14%7D%5C%20s)
Acceleration is given by
![a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{}{1.7\times 10^{-27}}\\\Rightarrow a=\ m/s^2](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7BF%7D%7Bm%7D%5C%5C%5CRightarrow%20a%3D%5Cdfrac%7B%3C-1.12%5Ctimes%2010%5E%7B-11%7D%2C%200%2C%200%3E%7D%7B1.7%5Ctimes%2010%5E%7B-27%7D%7D%5C%5C%5CRightarrow%20a%3D%3C-6.58824%5Ctimes%2010%5E%7B15%7D%2C%200%2C%200%3E%5C%20m%2Fs%5E2)
![v=u+at\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\\\Rightarrow v=\ m/s](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%20v%3D%3C3200%2C%20800%2C%200%3E%2B%3C-6.58824%5Ctimes%2010%5E%7B15%7D%2C%200%2C%200%3E%5Ctimes%202%5Ctimes%2010%5E%7B-14%7D%5C%5C%5CRightarrow%20v%3D%3C3200%2C%20800%2C%200%3E%2B%3C-6.58824%5Ctimes%2010%5E%7B15%7D%2C%200%2C%200%3E%5Ctimes%202%5Ctimes%2010%5E%7B-14%7D%5C%5C%5CRightarrow%20v%3D%3C3200%2C%20800%2C%200%3E%2B%3C-131.7648%2C%200%2C%200%3E%5C%5C%5CRightarrow%20v%3D%3C3068.2352%2C%20800%2C%200%3E%5C%20m%2Fs)
The velocity of the proton is ![\ m/s](https://tex.z-dn.net/?f=%3C3068.2352%2C%20800%2C%200%3E%5C%20m%2Fs)
Force = mass × acceleration
To find acceleration, we can divide the speed by the time it took:
acceleration = 2.40×10^7 / 1.8×10^-9
acceleration = 1.33×10^16
the mass is equal to the mass of an electron
force = (9.11×10^-31)(1.33×10^16)
force = 1.21×10^-14 N
Speed of light
According to Einstein, the speed of light is constant in all points of reference. In addition, he pointed out the speed of light is the maximum speed known since in practice one can never catch up with the beam of light. This is explained by his theory of relativity.
Answer:
Explanation:
a ) Slit separation d = .1 x 10⁻³ m
Screen distance D = 4 m
wave length of light λ = 650 x 10⁻⁹ m
Width of central fringe = λ D / d
= ![\frac{650\times10^{-9}\times4}{.1\times10^{-3}}](https://tex.z-dn.net/?f=%5Cfrac%7B650%5Ctimes10%5E%7B-9%7D%5Ctimes4%7D%7B.1%5Ctimes10%5E%7B-3%7D%7D)
= 26 mm
b ) Distance between 1 st and 2 nd bright fringe will be equal to width of dark fringe which will also be equal to 26 mm
c ) Angular separation between the central maximum and 1 st order maximum will be equal to angular width of fringe which is equal to
λ / d
= ![\frac{650\times10^{-9}}{.1\times10^{-3}}](https://tex.z-dn.net/?f=%5Cfrac%7B650%5Ctimes10%5E%7B-9%7D%7D%7B.1%5Ctimes10%5E%7B-3%7D%7D)
= 6.5 x 10⁻³ radian.