Online Test — Dual Nature of Matter and Radiation
18 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 18
The minimum energy needed to free an electron from a metal is the:
kinetic energy
work function
stopping potential
ionisation energy
Explanation: The work function $\phi_0$ is the minimum energy to eject an electron from the metal surface.
Question 2 of 18
Photoelectric current is directly proportional to the:
frequency
wavelength
intensity of light
work function
Explanation: Higher intensity supplies more photons per second, ejecting more electrons, so the current rises.
Question 3 of 18
The energy of a photon is given by:
$h\nu$
$\dfrac{h}{\nu}$
$\dfrac{\nu}{h}$
$mc$
Explanation: A photon carries energy $E = h\nu$.
Question 4 of 18
Einstein's photoelectric equation is:
$K_{max} = h\nu + \phi_0$
$K_{max} = h\nu - \phi_0$
$K_{max} = \phi_0 - h\nu$
$K_{max} = h\nu_0$
Explanation: Photon energy splits into the work function and the kinetic energy: $K_{max} = h\nu - \phi_0$.
Question 5 of 18
Below the threshold frequency, the number of photoelectrons emitted is:
maximum
small
zero
doubled
Explanation: If $\nu < \nu_0$ no electrons are emitted, whatever the intensity.
Question 6 of 18
The de Broglie wavelength of a particle of momentum $p$ is:
$hp$
$\dfrac{p}{h}$
$\dfrac{h}{p}$
$\dfrac{h}{2p}$
Explanation: $\lambda = \dfrac{h}{p}$ for every moving particle.
Question 7 of 18
The rest mass of a photon is:
equal to electron mass
infinite
zero
negative
Explanation: A photon has zero rest mass but still carries momentum $h/\lambda$.
Question 8 of 18
The maximum kinetic energy of photoelectrons depends only on the:
intensity
frequency
plate area
exposure time
Explanation: $K_{max} = h\nu - \phi_0$ depends on frequency (and the metal), not intensity.
Question 9 of 18
The stopping potential $V_0$ satisfies:
$K_{max} = eV_0$
$K_{max} = V_0/e$
$K_{max} = e/V_0$
$K_{max} = eV_0^2$
Explanation: The retarding voltage that just stops the fastest electron gives $eV_0 = K_{max}$.
Question 10 of 18
The slope of the stopping potential versus frequency graph equals:
$h$
$\dfrac{h}{e}$
$\dfrac{e}{h}$
$\phi_0$
Explanation: From $V_0 = \dfrac{h}{e}\nu - \dfrac{\phi_0}{e}$, the slope is $\dfrac{h}{e}$.
Question 11 of 18
The de Broglie wavelength of an electron accelerated through $V$ volts is about:
$\dfrac{12.27}{V}\,\text{\u00c5}$
$\dfrac{12.27}{\sqrt{V}}\,\text{\u00c5}$
$12.27\sqrt{V}\,\text{\u00c5}$
$\dfrac{V}{12.27}\,\text{\u00c5}$
Explanation: From $\lambda = h/\sqrt{2meV}$ the numerical formula is $\lambda = \dfrac{12.27}{\sqrt{V}}\,\text{\u00c5}$.
Question 12 of 18
The Davisson-Germer experiment confirmed the wave nature of:
photons
protons
electrons
neutrons
Explanation: Electron diffraction off a nickel crystal verified the de Broglie hypothesis for electrons.
Question 13 of 18
The photoelectric effect was first observed by:
Einstein
Hertz
de Broglie
Bohr
Explanation: Hertz noticed easier sparking under ultraviolet light in 1887.
Question 14 of 18
A photon of wavelength $\lambda$ has momentum:
$\dfrac{h}{\lambda}$
$h\lambda$
$\dfrac{\lambda}{h}$
$\dfrac{hc}{\lambda}$
Explanation: $p = \dfrac{h}{\lambda}$ for a photon.
Question 15 of 18
If the frequency of incident light is doubled (above threshold), the photon energy:
halves
stays same
doubles
becomes zero
Explanation: Since $E = h\nu$, doubling $\nu$ doubles the photon energy.
Question 16 of 18
The de Broglie wavelength in terms of kinetic energy $K$ is:
$\dfrac{h}{2mK}$
$\dfrac{h}{\sqrt{2mK}}$
$h\sqrt{2mK}$
$\dfrac{\sqrt{2mK}}{h}$
Explanation: With $p = \sqrt{2mK}$, $\lambda = \dfrac{h}{\sqrt{2mK}}$.
Question 17 of 18
Photoelectric emission is:
always delayed by minutes
instantaneous
delayed for weak light
only thermal
Explanation: Emission begins within about $10^{-9}\,\text{s}$ of illumination.
Question 18 of 18
The threshold frequency $\nu_0$ is related to the work function by:
$\phi_0 = h/\nu_0$
$\phi_0 = h\nu_0$
$\phi_0 = \nu_0/h$
$\phi_0 = h\nu_0^2$
Explanation: At threshold the photon energy just equals the work function: $\phi_0 = h\nu_0$.