Special Relativity

I was so bored on a flight, so I borrowed a pen from my neighbor passenger and wrote a quick derivation of a few results in special relativity:

Suppose an observer is doing a light speed measurement on a train by measuring the time it takes to go upwards and then reflect back and measure the time elapsed, he measures the time elapsed to be $2t$ and the distance traveled to be $2h$.

But the train is moving, so to an observer outside of the train, he measures the light moved in diagonal manner, let the time spent to be $t_0$, and the overall distance the light traveled is $2\sqrt{(vt_0)^2 + h^2}$. We assumed the two observers measure the same $h$ and the same $v$.

Both measurements are speed of light, so they equals:

$\frac{2h}{2t} = \frac{2\sqrt{(vt_0)^2 + h^2}}{2t_0} = c$

First, we cancel out all the $2$.

$\frac{h}{t} = \frac{\sqrt{(vt_0)^2 + h^2}}{t_0} = c$

Square the second equality to get

$(vt_0)^2 + h^2 = c^2t_0^2$

Rearranging, get 

$h^2 = (c^2 - v^2)t_0^2$

Dividing by $c^2$ get

$\frac{h^2}{c^2} = (1 - \frac{v^2}{c^2})t_0^2$

But $\frac{h^2}{c^2}$ is simply $t^2$ by the first equality, therefore

$t^2 = (1 - \frac{v^2}{c^2})t_0^2$

or 

$t = \sqrt{1 - \frac{v^2}{c^2}} t_0$.

The equation shows a few things:

First, $v \leq c$, for it does not make sense to have imaginary time. That indicates nothing can run faster than the speed of light.

Second, $0 \leq \sqrt{1 - \frac{v^2}{c^2}} \leq 1$, therefore, the time interval between the same events is shorter for the observers on the train. Suppose $t_0 = 1$ second, then $t$ is less than a second, so if there is a clock there, it doesn't tick yet. So for the observer outside the train, it appears that a moving clock go slower. This is called time dilation.

Third, since the both observers agrees on the relative velocity, the distance traveled, measured by the two observers are $vt$ and $vt_0$ respectively. We know $t \leq t_0$, therefore, to the observer on the train, distance appears shorter as well. This is called length contraction.

Chemistry (2)

Problem:

Solution:

The electron configuration of potassium is $1s^22s^22p^63s^23p^64s^1$, therefore the unpaired electron is occupying the 4s orbital.

Chemistry (1)

Problem:

Solution:

Oxygen has 8 protons always
The isotope with mass number 16 has 8 neutron

The isotope with mass number 17 has 9 neutron

The isotope with mass number 18 has 10 neutron