To find the speed of the electron, use the following formula:

\(\displaystyle v=\frac{2\pi ke^2}{nh}\)

\(\displaystyle k=\textup{Coulomb's constant}=8.99* 10^9 \frac{N \cdot m^2}{C^2}\)

\(\displaystyle n=\textup{first orbit}=1\)

\(\displaystyle h=\textup{Planks's constant}= 6.63 * 10^{-34}J\cdot s\)

\(\displaystyle e= \textup{electron charge}= 1.6 * 10^{-19}C\)

Substitute all the knowns and solve for velocity.

\(\displaystyle v=\frac{2\pi ke^2}{nh}=\frac{2\pi (8.99 * 10^9)(1.6 * 10^{-19})^2}{(1)(6.63 * 10^{-34})}\)

\(\displaystyle v=2.181 * 10^6 \frac{m}{s}\)

To answer this question, we'll need to utilize the equation that specifies the energy level of electrons within a hydrogen atom.

\(\displaystyle E=-\frac{13.6eV}{n^{2}}\)

Where \(\displaystyle n\) is equal to the electron energy level within the hydrogen atom. Also notice that this equation has a negative sign. This is because in its ground state, an electron is closest to the positively charged nucleus and thus has the lowest energy. As the energy level increases, the electron moves further and further away from the nucleus, thus gaining increasing energy. At an infinitely far away energy level, the electron will have a maximum energy value of zero. To find the difference between the second and fourth energy levels, we'll simply use the above equation for different values of \(\displaystyle n\).

\(\displaystyle \Delta E=-13.6eV\left ( \frac{1}{2^{2}}-\frac{1}{4^{2}} \right )\)

\(\displaystyle \Delta E=-13.6eV(0.1875)=-2.55eV\)

The negative sign for the change in energy just means that energy is being released in this process. We can drop the negative because we know that energy is being released.

\(\displaystyle 2.55eV*\frac{1.6* 10^{-19 J}}{1 eV}=4.08* 10^{-19} J\)

Now that we've found how much energy is contained in the released photon, we'll need to calculate its wavelength.

\(\displaystyle \Delta E=\frac{hc}{\lambda }\)

\(\displaystyle \lambda =\frac{hc}{\Delta E}\)

\(\displaystyle \lambda=\frac{(6.63*t 10^{-34}J\cdot s)(3.0*10^{8} \frac{m}{s})}{4.08* 10^{-19}J}\)

\(\displaystyle \lambda =4.875* 10^{-7}m=487.5nm\)

The incoming electron will lose kinetic energy during the collision, transfering this energy to the potential energy of the bound electron in the atom. Conservation of energy can be used to solve this problem. The general statement that energy is conserved is

\(\displaystyle K_{i}+U_{i}=K_{f}+U_{f}\)

where \(\displaystyle K\) is the kinetic energy and \(\displaystyle U\) is the potential energy. The incoming electron has kinetic energy and no potential energy. We are defining the initial state of the bound electron to be at \(\displaystyle 0eV\) so the total initial potential energy of the system is zero.

The incoming electron will still have kinetic energy after the collision but the bound electron will not since it is not a free electron. This means that

\(\displaystyle U_{f}=K_{i}-K_{f}\)

where

\(\displaystyle K=\frac{1}{2}m v^{2}\)

plugging this in - 

\(\displaystyle U_{f}=\frac{1}{2}m \left ( v_{f}\right )^{2}-\frac{1}{2}m \left ( v_{i}\right )^{2}=\frac{1}{2}m \left [ \left ( v_{f}\right )^{2}-\left ( v_{i}\right )^{2} \right ]\)

\(\displaystyle m\) is the mass of the electron. Plugging everything in and converting to \(\displaystyle eV\) gives

\(\displaystyle U_{f}=6.413\times10^{-19}J\left ( \frac{1eV}{1.6\times 10^{-19}J} \right )=4.01eV\)

During a energy level change in a hydrogen atom, the amount of energy either lost of gained is given by the following equation with respect to the initial and final energy levels shown below. 

\(\displaystyle E_n=\frac{-13.6eV}{n^2}\)

Recall that whenever electrons drop from higher energy levels to lower ones, energy can be released in the form of a photon. To obtain the amount of energy released, we mst take the difference in energy of the electrons at the particular energy levels:

\(\displaystyle \Delta E = -13.6eV\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )\)

\(\displaystyle \Delta E=-13.6eV\left(\frac{1}{2^2}-\frac{1}{4^2} \right )=-2.55eV\)

It is important to note that the negative energy difference corresponds to how much energy the photon is "taking away" as it leaves. Therefore, the photon leaves the atom with \(\displaystyle +2.55 J\) of energy.

Using

\(\displaystyle \Delta E=-2.18*10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\)

Plugging in values:

\(\displaystyle \Delta E=-2.18*10^{-18}(\frac{1}{1^2}-\frac{1}{3^2})\)

This will be the change in energy of the electron, which is the negative of the energy of the photon released.

\(\displaystyle \Delta E=-1.94*10^{-18}J\)

Thus, the energy of the photon is

\(\displaystyle E=1.94*10^{-18}J\)

Using the formula for the energy of an electron in a hydrogen atom's nth energy level:

\(\displaystyle E_n=\frac{-13.6}{n^2}eV\)

Plug in \(\displaystyle n=1\) and \(\displaystyle n=4\) then find the difference:

\(\displaystyle \Delta E=\frac{-13.6}{1}-\frac{-13.6}{16}\)

\(\displaystyle \Delta E=-12.75eV\)

Convert electronvolts to Joules:

\(\displaystyle \Delta E=-12.75eV*\frac{1.602*10^{-19}J}{1eV}\)

\(\displaystyle \Delta E=2.04*10^{-18} J\)

Using the following equation for the energy of an electron in Joules:

\(\displaystyle \Delta E=-2.18*10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\)

And

\(\displaystyle 1 \textup{mol}=6.02*10^{23}\text{molecules}\)

Combining equations and plugging in values:

\(\displaystyle \Delta E_{1mole}=6.02*10^{23}*-2.18*10^{-18}(\frac{1}{2^2}-\frac{1}{3^2})\)

\(\displaystyle \Delta E_{1mole}=-182kJ\)

\(\displaystyle 182kJ\) would be released

For this question, we need to compare the difference in energy levels of hydrogen atoms with electrons in different orbitals.

First, we will need to use the equation that describes the energy of an electron in a hydrogen atom.

\(\displaystyle E_{n}=\frac{-13.6\:eV}{n^{2}}\)

In the above expression, \(\displaystyle n\) represents the orbital in which the electron resides.

First, let's see what the electron energy level is in the ground state, which corresponds to \(\displaystyle n=1\).

\(\displaystyle E_{1}=\frac{-13.6\: eV}{1^{2}}=-13.6\: eV\)

Next, let's do the same thing for the \(\displaystyle n=3\) orbital.

\(\displaystyle E_{3}=\frac{-13.6\: eV}{3^{2}}=-13.6\: eV=-1.51\: eV\)

Next, we can find the difference in the energy values.

\(\displaystyle \Delta E=(-1.51\: eV)-(-13.6\: eV)=12.09\: eV\)

To answer this question, we'll need to consult the periodic table. From the table, we know that magnesium's atomic number (the number of protons it contains in its nucleus) is 12, and sodium's is 11. We also need to realize that the mass number for each (the number of protons plus neutrons contained in the nucleus) is the same. Since the mass numbers are the same but the atomic numbers differ by one, then we can infer that a neutron is undergoing a decay into a proton and a so called positron, \(\displaystyle \beta ^{+}\). The overall reaction is as follows:

\(\displaystyle _{12}^{23}\textrm{Mg}\rightarrow\: _{11}^{23}\textrm{Na}+e^{+}\)

Furthermore, it cannot be alpha decay, because in this process an alpha nucleus is released and the reactant's mass number and atomic number would both change. It also cannot be gamma decay, because in this process there is no change in atomic or mass numbers. Finally, it cannot be electron capture because in this process, an electron combines with a proton to generate a neutron. Thus, the mass number would not change, but the atomic number would increase by one. But in the question stem, we know the atomic number is decreasing by one rather than increasing.

The correct answer is quarks. Quarks usually have charges of \(\displaystyle \pm\frac{1}{3}e\) or \(\displaystyle \pm\frac{2}{3}e\). They are usually bound with other quark particles and could be mixed to form hadrons. Tau is part of the leptons family and has a charge of \(\displaystyle -1e\). Graviton does not have a charge, and is a hypothetical particle. The tachyon is a hypothetical particle assumed to be faster than light. Hadrons are strong composite particles that are composed of quarks and will result to a net integer charge.