QCAA Physics Alternative Sequence Quantum theory

A positively ionised metal would exert a stronger attractive force on the electrons, effectively increasing the energy required to remove them and decreasing their maximum kinetic energy.

According to the photoelectric equation $K_{\text{max}} = hf - \Phi$, using a metal with a larger work function ($\Phi$) would decrease the maximum kinetic energy of the ejected electrons.

Increasing the intensity of the incident light increases the number of ejected electrons per second, but it does not change the energy of individual photons or the maximum kinetic energy of the electrons.

Decreasing the wavelength of the incident light increases the energy of the incident photons ($E = \frac{hc}{\lambda}$), which directly increases the maximum kinetic energy of the ejected electrons according to $K_{\text{max}} = \frac{hc}{\lambda} - \Phi$.

This incorrect value is obtained by multiplying the energy by the speed of light ($E \cdot c$), rather than using the correct wave energy formula.

This incorrect value is obtained by dividing Planck's constant by the energy ($h/E$), which calculates the inverse of the frequency, not the wavelength.

Using the equation $E = \frac{hc}{\lambda}$, we can rearrange to find $\lambda = \frac{hc}{E}$. Plugging in the constants gives $\lambda = \frac{(6.626 \times 10^{-34} \text{ J}\cdot\text{s})(3.0 \times 10^8 \text{ m/s})}{2.4 \times 10^{-23} \text{ J}} \approx 8.3 \times 10^{-3} \text{ m}$.

This incorrect value is the inverse of the wavelength ($1/\lambda$), obtained by calculating $\frac{E}{hc}$ instead of $\frac{hc}{E}$.