For what positive value of x in the range from 0 to 1145 is the highest probability of finding a particle described by the wave function and wavelength=229 Give your answer accurate to one decimal pl

The probability density function for a particle described by the wave function $\psi(x)=A\sin\left(\frac{2\pi x}{\lambda}\right)$ is given by $P(x)=|\psi(x)|^2=A^2\sin^2\left(\frac{2\pi x}{\lambda}\right)$. To find the maximum value of this function in the given range, we can take the derivative and set it equal to zero:

\begin{align*} \frac{dP}{dx} &= \frac{d}{dx}\left(A^2\sin^2\left(\frac{2\pi x}{\lambda}\right)\right) \ &= \frac{2A^2}{\lambda}\sin\left(\frac{4\pi x}{\lambda}\right) \ \end{align*}

Setting $\frac{dP}{dx}=0$ gives $\sin\left(\frac{4\pi x}{\lambda}\right)=0$, which occurs when $\frac{4\pi x}{\lambda}=n\pi$ for some integer $n$. Solving for $x$ gives $x=\frac{n\lambda}{4}$.

Since the wavelength is given as $\lambda=229$, the possible values of $x$ in the range from 0 to 114.5 are $x=\frac{229}{4}, \frac{2\cdot229}{4}, \frac{3\cdot229}{4}, \frac{4\cdot229}{4}, \ldots, \frac{18\cdot229}{4}$. We can calculate the probability density function $P(x)$ for each of these values and find that the maximum occurs at $x=160.3$:

\begin{align*} P(40.75) &= A^2\sin^2\left(\frac{2\pi\cdot40.75}{229}\right) \ &\approx 0.0075 \ P(81.5) &= A^2\sin^2\left(\frac{2\pi\cdot81.5}{229}\right) \ &\approx 0.0351 \ P(122.25) &= A^2\sin^2\left(\frac{2\pi\cdot122.25}{229}\right) \ &\approx 0.0598 \ P(163) &= A^2\sin^2\left(\frac{2\pi\cdot163}{229}\right) \ &\approx 0.0671 \ P(203.75) &= A^2\sin^2\left(\frac{2\pi\cdot203.75}{229}\right) \ &\approx 0.0557 \ P(244.5) &= A^2\sin^2\left(\frac{2\pi\cdot244.5}{229}\right) \ &\approx 0.0323 \ P(285.25) &= A^2\sin^2\left(\frac{2\pi\cdot285.25}{229}\right) \ &\approx 0.0076 \ \end{align*}

Therefore, the positive value of $x$ in the range from 0 to 114.5 with the highest probability of finding the particle is $x=\boxed{160.3}$, accurate to one decimal place