The graph below is obtained for an experiment on the photoelectric effect using different frequencies of light and a given metal plate: The threshold frequency for the metal is 6,8 x 10^14 Hz - NSC Physical Sciences - Question 10 - 2017 - Paper 1

The speed remains unchanged.

To determine if the photoelectric effect is observed, we first need to calculate the frequency of the light using the equation:
$f = \frac{c}{\lambda}$
where $c = 3 \times 10^8 \ m/s$ and $\lambda = 6 \times 10^{-7} \ m$.
Thus,
$f = \frac{3 \times 10^8}{6 \times 10^{-7}} = 5 \times 10^{14} \ Hz$
Since this frequency (5 x 10^14 Hz) is less than the threshold frequency (6.8 x 10^14 Hz), the photoelectric effect will NOT be observed.

The energy of the photon can be calculated using the equation:
$E_{photon} = hf = h \times 7.8 \times 10^{14} \ Hz$
where $h = 6.63 \times 10^{-34} \ J\cdot s$.
Substituting, we find:
$E_{photon} = (6.63 \times 10^{-34})(7.8 \times 10^{14}) = 5.17 \times 10^{-19} \ J$
The maximum kinetic energy of the photoelectron is given by:
$KE_{max} = E_{photon} - W$
where $W$ is the work function. Assuming $W < E_{photon}$, the kinetic energy can be equated to
$KE = \frac{1}{2} mv_{max}^2$
Solving for $v_{max}$, we get
$v_{max} = \sqrt{\frac{2 \cdot KE_{max}}{m}}$
Utilizing the known values, with a typical electron mass $m \approx 9.11 \times 10^{-31} \ kg$, we can find $v_{max}$.