- What is the net static electric charge on a metal sphere having an excess of +3 elementary charges?
- \(4.7 \times 10^{-18}\ \text{C}\)
- \(8.4 \times 10^{-19}\ \text{C}\)
- \(4.8 \times 10^{-19}\ \text{C}\)
- \(8.7 \times 10^{-19}\ \text{C}\)
Charge is quantized in units of the elementary charge:
\(e \approx 1.60\times 10^{-19}\ \text{C}\)
For an excess of \(+3e\):
\(Q = 3e \approx 3(1.60\times 10^{-19}) = 4.8\times 10^{-19}\ \text{C}\)
- A hydrogen atom could have an electron energy level transition from \(n=2\) to \(n=3\) by absorbing a photon having an energy of how many eV?
- 1.89
- 2.04
- 2.25
- 2.87
Hydrogen levels:
\(E_n = -\dfrac{13.6\ \text{eV}}{n^2}\)
Thus \(E_2=-3.40\ \text{eV}\) and \(E_3=-1.51\ \text{eV}\). The photon energy needed is:
\(\Delta E = E_3 – E_2\) \(\approx (-1.51)-(-3.40)\) \(=1.89\ \text{eV}\)
- Two solid metal blocks are placed in an insulated container. If there is a net flow of heat between the blocks, they must have different…
- Initial temperatures
- Specific heat values
- Melting points
- Heats of fusion
Heat flows spontaneously from higher to lower temperature until thermal equilibrium is reached (zeroth/second laws). Specific heat or melting point doesn’t determine the direction of heat flow; a temperature difference does.
- A p-type semiconductor is formed by adding impurities, which provide extra…
- Electrons
- Neutrons
- Photons
- Holes
Acceptors create vacant states in the valence band that act as the majority carriers in p-type material.
- A student measures a current of 0.05 amperes through a p-type semiconductor sample with ohmic contacts. If the battery connections are reversed, what will the current through the semiconductor be (for the same applied voltage magnitude)?
- Less than 0.05 amperes
- Greater than 0.05 amperes
- The same
For a uniform p-type sample with ohmic contacts (not a diode junction), \(I\propto V\) by Ohm’s law.
Reversing polarity just reverses the direction of hole drift. The magnitude stays \(0.05\ \text{A}\) for the same \(|V|\).
- What is the approximate binding energy of a helium nucleus that has a mass defect of \(5.2 \times 10^{-29}\ \text{kg}\)?
- \(4.6 \times 10^{-12}\ \text{J}\)
- \(4.6 \times 10^{-13}\ \text{J}\)
- \(4.7 \times 10^{-12}\ \text{J}\)
- \(4.7 \times 10^{-13}\ \text{J}\)
Use mass–energy equivalence:
\(E=\Delta m\,c^2\) \(\approx (5.2\times 10^{-29}\ \text{kg})(3.00\times 10^8\ \text{m/s})^2\) \(\approx 4.68\times 10^{-12}\ \text{J}\) \(\approx 4.7\times 10^{-12}\ \text{J}\)
That’s about \(2.9\times 10^7 \text{ eV} \approx 29 \text{ MeV}\), consistent with He-4.
- Which particle cannot be accelerated by a cyclotron?
- Proton
- Neutron
- Electron
- α-particle
A cyclotron accelerates charged particles using an RF electric field and bends them with a magnetic field. Neutrons are neutral and don’t gain energy from the accelerating electric field.
- A 96 gram sample of a radioactive nuclide is placed in a container. After 12 minutes, only 6 grams of the sample remain undecayed. What is the half-life of the nuclide?
- 3 minutes
- 4 minutes
- 5 minutes
- 6 minutes
Decay law:
\(N=N_0\left(\tfrac12\right)^{t/T_{1/2}}\)
Here, \(\tfrac{N}{N_0}=\tfrac{6}{96}=\tfrac{1}{16}=\left(\tfrac12\right)^4\), so \(\tfrac{t}{T_{1/2}}=4\).
With \(t=12 \text{ min}\):
\(T_{\tfrac{1}{2}}=\dfrac{12}{4}=3 \text{ min}\)
- What is the principal reason for using neutrons to bombard a nucleus?
- Neutrons have a relatively low atomic mass
- Neutrons can be easily accelerated
- Neutrons have a very high kinetic energy
- Neutrons are not repelled by the nucleus
Because neutrons are uncharged, they don’t face a Coulomb barrier when approaching a positively charged nucleus, making nuclear reactions more probable at lower energies.
- What is the magnitude of the gravitational force between an electron and a proton separated by a distance of \(1.0 \times 10^{-10}\ \text{m}\)?
- \(1.0 \times 10^{-43}\ \text{N}\)
- \(1.0 \times 10^{-47}\ \text{N}\)
- \(2.0 \times 10^{-43}\ \text{N}\)
- \(3.05 \times 10^{-47}\ \text{N}\)
At atomic separations, the gravitational pull is given by Newton’s law, \(F=G\tfrac{m_e m_p}{r^2}\).
Using \(G=6.67\times10^{-11}\), \(m_e=9.11\times10^{-31}\ \text{kg}\), \(m_p=1.67\times10^{-27}\ \text{kg}\), and \(r=1.0\times10^{-10}\ \text{m}\), we find \(m_e m_p\approx1.52\times10^{-57}\) and \(r^2=1.0\times10^{-20}\), giving \(F\approx1.0\times10^{-47}\ \text{N}\).
This reflects how much weaker gravity is than electric forces at atomic scales.
- A sphere has a net excess charge of \(-4.8 \times 10^{-19}\ \text{C}\). The sphere must have an excess of…
- 1 electron
- 1 proton
- 3 electrons
- 3 protons
\(n=\dfrac{|Q|}{e}\) \(=\dfrac{4.8\times 10^{-19}}{1.60\times 10^{-19}}\) \(=3\)
The negative sign indicates excess electrons, not protons.
- An excited hydrogen atom returns to its ground state. Which of the following is a possible energy change for the atom?
- Loss of 10.20 eV
- Gain of 10.20 eV
- Loss of 11.70 eV
- Gain of 11.70 eV
Returning to ground means the atom loses energy by emitting a photon. The \(n=2\to 1\) transition is 10.20 eV. Gains aren’t possible when de-exciting, and 11.70 eV doesn’t correspond to an allowed Lyman transition.
- What is conserved during a collision between a proton and an electron?
- Energy
- Momentum
- Energy and momentum
- Neither energy nor momentum
In an isolated system, total energy (including rest and kinetic) and total momentum are conserved in all collisions, elastic or inelastic.
- As the temperature of a surface increases, how does the rate of thermionic emission change?
- Electrons are emitted at a lower rate.
- Electrons are emitted at a higher rate.
- Protons are emitted at a lower rate.
- Protons are emitted at a higher rate.
Thermionic emission increases rapidly with temperature. The emitted particles are electrons, not protons.