Quantum and Nuclear Physics - AP Physics 2

Using

Converting to and plugging in values

Using

is the photon energy

is Plank's constant,

is the speed of light,

is the wavelength

Converting to and plugging in values:

The wavelength of the light beam doesn't depend on the speed through the medium. As such, as can solve for wavelength by simply solving:

Where is frequency and is the speed of light in a vacuum.

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

Where 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 .

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.

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

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

where is the kinetic energy and 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 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

where

plugging this in -

is the mass of the electron. Plugging everything in and converting to gives

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.

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:

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 of energy.

Using

Plugging in values:

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

Thus, the energy of the photon is

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

Plug in and then find the difference:

Convert electronvolts to Joules:

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

And

Combining equations and plugging in values:

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.

In the above expression, 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 .

Next, let's do the same thing for the orbital.

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

Fusion is the process of combining two or more atoms to form a larger atom. To remember this, think of how welders fuse metals together. (Though the term is the same, they aren't actually the same thing; this is just to help you remember.). Fusion is a very energetic reaction that takes place in high-heat, high-pressure environments, like the inside of stars. Fusion releases lots of energy, which is why stars are so energetic.

Fission is the process of splitting a signle atom into multiple atoms. It doesn't normally occur in nature, though some super heavy elements, like plutonium, can be spontaneously fissile, which means they can undergo fission seemingly at random. This is a rare thing for an element to do, which is why it's said that fission doesn't normally occur in nature.

This is an example of one of Einstein's greatest ideas: the relation between the mass of an object/particle, and the energy contained by the mass. This is given as

In order to calculate the energy in our particle, we must make sure that the mass is in units of .

Now we can plug in numbers to our equation and solve for energy.

In this question, we're presented with information concerning the mass of a uranium atom. We're told two values: the mass of a uranium atom as measured, and the predicted mass of a uranium atom. We're then asked to determine the nuclear binding energy for uranium.

In order to solve this question, we have to realize the significance of the discrepancy between the observed and predicted mass of uranium. The predicted mass is calculated by adding up the individual masses of each constituent proton, electron, and neutron. However, the reason why the measured mass is less than the predicted mass is due to energy-mass equivalence. When the constituent protons and neutrons come together to form the nucleus, some of their mass is converted into energy, and it is this energy which holds these constituent nucleons together. Because some of the mass is converted into energy, the observed mass is less than what we would predict.

Now that we understand why there is a discrepancy between observed and predicted mass, we can calculate the nuclear binding energy by using Einstein's famous equation.

This equation states that the nuclear binding energy is equal to the difference between observed and predicted mass, multiplied by the speed of light squared. So to solve for energy, we can plug in the values given to us.

The energy from the two grams of helium can be found using

This energy can then be equated to the man's kinetic energy, which can then be used to find the man's velocity.

In this question, we're given the mass defect of an atom's nucleus and are asked to find the binding energy for this atom.

To begin with, it's important to understand that when protons and neutrons come to be held together within the nucleus of an atom, there is a tremendously powerful force holding them together. This incredibly large force accounts for the mass defect. In other words, the total mass of the nucleus is smaller than the sum of the individual masses of the protons and neutrons that make up that nucleus, and this is due to the strong force.

Einstein's mass-energy equivalence explains the observable mass defect; the mass lost is converted into an enormous amount of energy according to the following equation.

But first, we'll need to convert the mass given to us in the question stem into grams.

Furthermore, because we know the value for the speed of light, we can use this information to solve for the binding energy.

First of all, the intermediate nuclear force isn't a real force.

Gravity is not responsible for this, because on the scale of quantum mechanical phenomena, gravity has negligible effect, and can be disregarded.

The electromagnetic force doesn't hold the nucleus together, and is actually trying to rip it apart, due to the fact that like charges repel and the nucleus is full of like charges (protons). Accordingly, the force that actually is responsible for holding it together must necessarily be significantly more powerful compared to the electromagnetic force to resist the intrinsic repelling the protons have towards each other.

The weak nuclear force operates on leptons and quarks, and is involved in many of the radioactive decays in nuclear physics, such as beta decay, where a proton decays into a neutron, where it was first revealed.

Since the other three valid forces aren't responsible, that leaves the strong nuclear force. It is the strongest of the four fundamental forces, as it prevents protons from flying away from each other due to their proximity and charge. The strong force mediates over the quarks that make up the protons and neutrons.

In this question, we're told that an atom undergoes a series of decays. We're then asked to determine how the atomic number of that atom has changed.

Let's look at the first type of decay, alpha decay. During alpha decay, the atom emits a helium nucleus, which consists of two protons and two neutrons. Thus, for each alpha decay, the atom will lose two protons. So two alpha decays would result in a net loss of four protons.

Next, let's look at positron decay. In this type of decay, a proton is converted into a positron and a neutron. The neutron stays in the atoms's nucleus, while the positron is emitted. Thus, positron decay results in a loss of one proton. Consequently, two positron decays result in a total loss of two protons.

Finally, gamma decay does not cause a change in the atom's atomic number or mass number. Gamma decay simply releases energy.

So, in total, we have four protons lost from alpha decays and two protons lost from positron decays. Thus, there is a total loss of six protons, corresponding to a decrease in the atomic number by six.

The energy of the quantum system in the state is given by

where is Planck's constant, is the mass of the proton and is the length of the box. The energy of a photon is given by

where is the frequency, is the speed of light and is the wavelength. Setting these equal we can solve for ,

Since the ground state is , the proton must be in the second excited state.

From a statistical standpoint, the expectation value of the position, , can only tell us the most probable location of the particle. A central idea in quantum mechanics is that we can never really know exactly where a particle is as a function of time, but rather where we are most likely to find the particle if we choose to observe it.