NESA Physics Light: Quantum Model

This incorrect value results from simply multiplying the energy by the speed of light ($E \times c$), ignoring Planck's constant and the correct formula.

This answer is obtained by dividing Planck's constant by the energy ($h/E$) but neglecting to multiply by the speed of light ($c$).

This value represents the wavenumber ($1/\lambda \approx 120$ m$^{-1}$), which is the reciprocal of the wavelength rather than the wavelength itself.

This value results from incorrectly dividing the temperature by Wien's constant ($T/b$) without involving the speed of light, which does not yield a frequency.

This value represents the peak wavelength in meters ($4.8 \times 10^{-7} \text{ m}$), not the frequency. You likely forgot to convert wavelength to frequency using $f = c/\lambda$.

This value is the period of the wave ($T = 1/f$), which represents the time for one cycle, rather than the frequency.

Using a positively ionised metal would increase the electrostatic attraction on the electrons, effectively increasing the energy required to escape and decreasing the kinetic energy.

According to the photoelectric equation $K_{max} = h\nu - \Phi$, increasing the work function ($\Phi$) reduces the remaining energy available for kinetic energy.

Increasing intensity increases the number of photons and ejected electrons (photocurrent), but does not change the energy of individual photons or the maximum kinetic energy of the electrons.

White light passing through a prism demonstrates dispersion and refraction, which are phenomena explained by the wave nature of light and its varying wavelengths.

Diffraction is the bending of light as it passes through a slit, which is a classic characteristic of waves rather than particles.

Interference patterns are created by the superposition of overlapping waves, providing foundational evidence for the wave model of light.

This option is incorrect. It appears to confuse the numerical value of the frequency ($1.7$) with the final energy, ignoring Planck's constant and the necessary unit conversions.

This option represents the kinetic energy in Joules ($1.3 \times 10^{-19}$ J), not electron-volts. To obtain the correct answer in eV, you must divide this value by the elementary charge ($1.60 \times 10^{-19}$ J/eV).

This value is extremely small and physically unrealistic for the kinetic energy of a photoelectron in this context. It likely results from an arithmetic error or incorrect formula application.

The spectral distribution of black-body radiation is a continuous curve with a single peak intensity wavelength determined by the temperature, not two peaks.

This describes a misunderstood version of the photoelectric effect; a black body is defined by its perfect absorption and thermal emission of electromagnetic radiation, not by electron emission.

According to Wien's Displacement Law, the peak wavelength of emission is inversely proportional to temperature ($\lambda_{\text{max}} \propto 1/T$), meaning the peak shifts to shorter wavelengths as the object gets hotter.

This is incorrect because Experiment 2 (the photoelectric effect) demonstrates particle-like behavior that cannot be explained by the wave theory of light.

This is incorrect because the bending of light (diffraction) observed in Experiment 1 is a characteristic behavior of waves, not particles.

This is incorrect because the observations provided do not address travel in a vacuum, and light travels through a vacuum regardless of the model used to describe it.

Incorrect. The energy supplied by the incident photon is represented by the term $hf$, not $\phi$.

Incorrect. The energy retained by the electron as kinetic energy after being ejected is represented by $K_{\text{max}}$.

Incorrect. The energy left over after freeing the electron becomes its maximum kinetic energy, represented by $K_{\text{max}}$.

This value represents the threshold frequency calculated using only the work function ($\nu = \Phi/h$), ignoring the additional energy required for the electron's kinetic velocity.

This incorrect answer results from subtracting the work function energy from the kinetic energy (or vice versa) instead of adding them to determine the total photon energy.

This calculation incorrectly equates the photon energy solely to the electron's kinetic energy ($\nu = KE/h$), neglecting the work function energy required to eject the electron from the metal.

In the photon model, increasing light intensity increases the number of photons (and thus the number of ejected electrons), but it does not change the energy of individual photons or the maximum kinetic energy of the electrons.

The maximum kinetic energy of ejected electrons depends on the metal's work function ($K_{max} = hf - \Phi$), which is a property that varies depending on the specific type of metal used.

According to the photon model, if the light's wavelength is above the threshold, no electrons will be emitted regardless of how long the metal is exposed to the light, because individual photons lack the energy to eject an electron.