Test: Atomic Nucleus - MCAT MCQ

Beta minus decay results in the release of an electron
Beta minus decay converts a neutron into a proton
The atomic number, and thus the identity of the element, increases by one during beta minus decay
The beta decay product of M, g is thus .

• In ordinary fission, gamma ray photons are not emitted by a decaying nucleus.
• The reason that fission is energetically favorable is that the sum of the binding energies of the daughter nuclei is less than that of the parent nuclei.
• During ordinary fission, the electronic energy levels do not strongly play a role in the decay process.
• In ordinary fission, the binding energy of the parent nucleus is transformed into the kinetic energy of the daughter nuclei.

The mass of each isotope is given by its superscript.
The atomic mass of an element is the average of all isotopes of an element, weighted by their natural abundance.
Because nitrogen-14 is much more common than nitrogen-12, the average atomic mass of the element must be closer to that of nitrogen-14 than of nitrogen-12.
The atomic mass is thus .7 × 14 + .3 ∗ 12 = 13.4 amu.

Note that the semilog axis is given in base ten logarithm, and not base 1/2
You can solve this problem by approximating, without using the change-of-basis formula for logarithms
Looking carefully at the y-axis, It’s apparent that the 1000 kg mass drops down to 500 kg in way less than 60 years, but not quite as abruptly as 1 yr.
The half-life of the isotope is 20 years.

The half-life is an approximate quantity, the question requires estimation based on what is known about the average rates of radioactive decay via an exponential law.

After 10 seconds, there are, on average, only 32 atoms. Repeating this division (or observing that 2⁶ = 64), it is apparent that we would only expect there to be one atom after 10∗ 6 = 60 s.

Thus after an additional 10 seconds (a total of 7 half-lives) we would expect there to be only half an atom; because we can’t have half an atom we would estimate there to be none of the isotope left at this time

In the standard derivation of the mass spectrometer, the centripetal acceleration due to the Lorentz force results in an equation for the radius of deflection, r = mv/qB, where q and m are the charge and mass of the isotope, respectively.  B is the deflecting field, which is the same for all isotopes.  v is the velocity with which the particles enter the magnetic field region.

The voltage that accelerates electrons to velocity v before entering the field region is set by energy conservation, q/V = 1/2 mv² where V is a voltage set by the machine that remains constant for all ions. This means that the entering velocity v depends only on the ratio q/m, for the isotopes.

The pair of isotopes, are thus impossible to distinguish because they have the same charge-to-mass ratio and thus are deflected the same amount by the magnetic field.

The Lorentz force law states that the magnitude of the force acting on the isotopes goes as qvB sin θ, where θ is the angle between the field lines and the velocity.

The velocity of the particle can be written in terms of a component perpendicular to the field, v⊥​ and a component along the field line, v||

The Lorentz force in the perpendicular direction is thus qv⊥ ​B, but it is zero in the parallel direction. The particle thus only undergoes circular motion in the plane perpendicular to the field line, and it travels inertially along the direction of the field lines.

The charged particle thus traces out a corkscrew trajectory as it moves in the device.

Several answer choices can be eliminated by considering the properties of the exponential function. The inverse of the decay rate is 10 s, which means that the amount should decrease by several factors of e in 30 s

First write the exponential decay equation, N(t) = N₀e −λt and identify that N₀ = 10 Kg and λ = 0.1 1/s

Inserting t = 30 s we find that N = 10 × e^-3 ≈ .5 kg

All nuclear reactions obey conservation of mass
The mass of an isotope is given by its superscript
Each lone neutron has a mass of 1
The total mass of the released isotopes and neutrons must equal 236
Because 92 +141 + 3 = 236, the only valid isotope is 

Recall what types of particles are released during each process: Alpha decay releases helium nuclei, beta decay releases electrons, gamma decay releases photons, and neutron decay releases neutrons.

An alpha particle or helium nucleus consists of 2 neutrons and 2 protons, for a total mass of 4 atomic mass units.

Alpha decay results in the largest mass change for a decaying nucleus