Online Test — Atoms and Nuclei
20 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 20
Rutherford's nuclear model of the atom failed to explain:
the existence of the nucleus
the stability of the atom
the deflection of $\alpha$ particles
the mass of the atom
Explanation: An accelerating orbital electron would radiate and spiral in, so the model could not explain atomic stability.
Question 2 of 20
Bohr's quantisation condition for angular momentum is:
$mvr = \frac{nh}{4\pi}$
$mvr = \frac{2\pi}{nh}$
$mvr = nh$
$mvr = \frac{nh}{2\pi}$
Explanation: Angular momentum is quantised as $mvr = \frac{nh}{2\pi}$.
Question 3 of 20
The energy of the electron in the ground state of hydrogen is:
$-13.6\ \text{eV}$
$-3.4\ \text{eV}$
$-1.51\ \text{eV}$
$0$
Explanation: $E_1 = -\frac{13.6}{1^2} = -13.6\ \text{eV}$.
Question 4 of 20
The radius of the $n$th Bohr orbit is proportional to:
$\frac{1}{n}$
$n^2$
$n$
$\frac{1}{n^2}$
Explanation: $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2} \propto n^2$.
Question 5 of 20
Spectral lines of hydrogen ending on $n_1 = 2$ form the:
Lyman series
Paschen series
Balmer series
Brackett series
Explanation: Transitions to $n_1 = 2$ give the Balmer series (visible).
Question 6 of 20
In the Rydberg formula, $R$ has the value approximately:
$1.097 \times 10^{5}\ \text{m}^{-1}$
$1.097 \times 10^{9}\ \text{m}^{-1}$
$6.6 \times 10^{-34}\ \text{m}^{-1}$
$1.097 \times 10^{7}\ \text{m}^{-1}$
Explanation: The Rydberg constant is $R = 1.097 \times 10^7\ \text{m}^{-1}$.
Question 7 of 20
The number of neutrons in ${}^{56}_{26}\text{Fe}$ is:
30
26
56
82
Explanation: $N = A - Z = 56 - 26 = 30$.
Question 8 of 20
Nuclei with the same atomic number but different mass numbers are:
isobars
isotopes
isotones
isomers
Explanation: Same $Z$, different $A$ defines isotopes.
Question 9 of 20
The nuclear radius depends on the mass number as:
$R = R_0 A^{1/2}$
$R = R_0 A^2$
$R = R_0 A^{1/3}$
$R = R_0 A$
Explanation: $R = R_0 A^{1/3}$, $R_0 = 1.2\ \text{fm}$.
Question 10 of 20
One atomic mass unit (u) is equivalent to:
$931.5\ \text{eV}$
$931.5\ \text{keV}$
$931.5\ \text{GeV}$
$931.5\ \text{MeV}$
Explanation: $1\ \text{u} = 931.5\ \text{MeV}/c^2$.
Question 11 of 20
The mass defect of a nucleus is defined as:
mass of nucleons minus mass of nucleus
mass of nucleus minus mass of nucleons
mass of protons only
half the nuclear mass
Explanation: $\Delta m = [Z m_p + (A - Z) m_n] - M_{\text{nucleus}}$.
Question 12 of 20
The binding energy per nucleon is maximum near:
uranium
iron
hydrogen
helium
Explanation: The curve peaks at about $8.8\ \text{MeV}$ near iron-56.
Question 13 of 20
Gamma rays are:
positively charged
negatively charged
high-energy photons (uncharged)
helium nuclei
Explanation: $\gamma$ rays are uncharged high-energy electromagnetic photons.
Question 14 of 20
In $\beta^-$ decay, what is emitted along with the electron?
a proton
a neutron
an alpha particle
an antineutrino
Explanation: A neutron becomes a proton, emitting an electron and an antineutrino $\bar{\nu}$.
Question 15 of 20
The radioactive decay law is:
$N = N_0 e^{-\lambda t}$
$N = N_0 e^{\lambda t}$
$N = N_0 \lambda t$
$N = \frac{N_0}{\lambda t}$
Explanation: $N = N_0 e^{-\lambda t}$ — exponential decay.
Question 16 of 20
After two half-lives, the fraction of nuclei remaining is:
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{8}$
$\frac{1}{3}$
Explanation: $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
Question 17 of 20
The mean life $\tau$ and decay constant $\lambda$ are related by:
$\tau = \lambda$
$\tau = 0.693\lambda$
$\tau = \frac{1}{\lambda}$
$\tau = \lambda^2$
Explanation: $\tau = \frac{1}{\lambda} = 1.44\,T_{1/2}$.
Question 18 of 20
The approximate energy released per fission of a ${}^{235}_{92}\text{U}$ nucleus is:
$2\ \text{MeV}$
$200\ \text{MeV}$
$2\ \text{GeV}$
$20\ \text{keV}$
Explanation: About $200\ \text{MeV}$ is released per uranium-235 fission.
Question 19 of 20
Nuclear density is:
greatest for heavy nuclei
least for light nuclei
nearly the same for all nuclei
zero for stable nuclei
Explanation: Since both mass and volume scale with $A$, density is constant ($\approx 2.3\times10^{17}\ \text{kg m}^{-3}$).
Question 20 of 20
The first line of the Balmer series ($\text{H}_\alpha$) has wavelength about:
$656\ \text{nm}$
$121\ \text{nm}$
$486\ \text{nm}$
$103\ \text{nm}$
Explanation: The $n=3 \to n=2$ transition gives $\lambda \approx 656\ \text{nm}$ (red).