IMO Practice Test — Dual Nature of Matter and Radiation
12 Questions • 15 min • Olympiad level
15:00
Question 1 of 12
The work function of a metal is $3.0\,\text{eV}$. Light of energy $5.0\,\text{eV}$ falls on it. The maximum kinetic energy of the photoelectrons is:
$8.0\,\text{eV}$
$2.0\,\text{eV}$
$1.5\,\text{eV}$
$3.0\,\text{eV}$
Explanation: $K_{max} = h\nu - \phi_0 = 5.0 - 3.0 = 2.0\,\text{eV}$.
Question 2 of 12
If the stopping potential is $2.0\,\text{V}$, the maximum kinetic energy of the photoelectrons is:
$2.0\,\text{eV}$
$1.0\,\text{eV}$
$0.5\,\text{eV}$
$4.0\,\text{eV}$
Explanation: $K_{max} = eV_0 = e(2.0\,\text{V}) = 2.0\,\text{eV}$.
Question 3 of 12
The threshold wavelength of a metal is $620\,\text{nm}$. Its work function (with $hc = 1240\,\text{eV nm}$) is:
$1.0\,\text{eV}$
$2.0\,\text{eV}$
$3.0\,\text{eV}$
$6.2\,\text{eV}$
Explanation: $\phi_0 = hc/\lambda_0 = 1240/620 = 2.0\,\text{eV}$.
Question 4 of 12
An electron accelerated through $150\,\text{V}$ has a de Broglie wavelength of about:
$1.0\,\text{\u00c5}$
$1.5\,\text{\u00c5}$
$0.5\,\text{\u00c5}$
$2.0\,\text{\u00c5}$
Explanation: $\lambda = 12.27/\sqrt{150} = 12.27/12.25 \approx 1.0\,\text{\u00c5}$.
Question 5 of 12
Two photons have wavelengths in the ratio $1:2$. The ratio of their energies is:
$1:2$
$2:1$
$1:4$
$4:1$
Explanation: $E = hc/\lambda$, so energy is inversely proportional to wavelength; ratio is $2:1$.
Question 6 of 12
If the intensity of incident light is doubled (frequency fixed, above threshold), the stopping potential:
doubles
halves
stays the same
becomes zero
Explanation: Stopping potential depends only on frequency, not intensity, so it is unchanged.
Question 7 of 12
A proton and an electron have the same kinetic energy. The one with the longer de Broglie wavelength is the:
proton
electron
both equal
depends on charge
Explanation: $\lambda = h/\sqrt{2mK}$; for equal $K$ the lighter electron has the longer wavelength.
Question 8 of 12
The maximum kinetic energy of photoelectrons is $1.0\,\text{eV}$ for light of $3.0\,\text{eV}$. The work function is:
$4.0\,\text{eV}$
$2.0\,\text{eV}$
$1.0\,\text{eV}$
$3.0\,\text{eV}$
Explanation: $\phi_0 = h\nu - K_{max} = 3.0 - 1.0 = 2.0\,\text{eV}$.
Question 9 of 12
Doubling the speed of an electron changes its de Broglie wavelength to:
double
half
four times
unchanged
Explanation: $\lambda = h/(mv)$ is inversely proportional to $v$, so it halves.
Question 10 of 12
The slope of a $V_0$ vs $\nu$ graph is the same for all metals because it equals:
$\phi_0$
$\nu_0$
$h/e$
$e/h$
Explanation: The slope is $h/e$, a universal constant independent of the metal.
Question 11 of 12
A particle of mass $m$ and kinetic energy $K$ has a de Broglie wavelength. If $K$ is increased four times, $\lambda$ becomes:
double
half
four times
one-quarter
Explanation: $\lambda \propto 1/\sqrt{K}$, so quadrupling $K$ halves the wavelength.
Question 12 of 12
In the Davisson-Germer experiment, the diffraction peak appears because electrons behave as:
pure particles
waves
photons
neutrons
Explanation: Diffraction is a wave property; the peak shows electrons act as matter waves.