Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The problem describes an interference pattern where the most intense fringe (peak) occurs at $x = 7.0 \; \rm{cm} $.
- There are slightly weaker fringes at $x = 6.0 \; \rm{cm} $ and $x = 8.0 \; \rm{cm} $.
- There are still weaker fringes at $x = 4.0 \; \rm{cm} $ and $x = 10.0 \; \rm{cm} $.
- There are two very weak fringes at $x = 1.0 \; \rm{cm} $ and $x = 13.0 \; \rm{cm} $.
- No electrons are detected for $x \lt 0 \; \rm{cm} $ or $x \gt 14 \; \rm{cm} $.
From all the above, we can draw the graph of $|\psi(x)|^2 $ as shown in the first figure below.
$$\color{blue}{\bf [b]}$$
- The wave function $\psi(x) $ is related to the square root of the probability density $|\psi(x)|^2 $.
- The graph of $\psi(x) $ will oscillate between positive and negative values, with its amplitude corresponding to the peaks in $|\psi(x)|^2 $.
- The highest amplitude will be at $x = 7.0 \; \rm{cm} $, with smaller oscillations at $x = 6.0 \; \rm{cm} $, $x = 8.0 \; \rm{cm} $, and weaker ones at other fringe locations.
From all the above, we can draw the graph of $|\psi(x)| $ as shown in the second figure below.
$$\color{blue}{\bf [c]}$$
The other possible graph for $\psi(x) $ is $-\psi(x) $ (the negative of it) since that will not affect the original function of $|\psi(x)|^2 $.
From all the above, we can draw the second version of the graph of $|\psi(x)| $ as shown in the third figure below (which is the same second graph but rotated 180$^\circ$).
