‎[tex]\psi(x) = A*(\frac{x}{x_{0}})^n*e^(\frac{-x}{x_{0}}) [/tex] Where A, n, and X0 are constants.

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HOME > > ‎[tex]\psi(x) = A*(\frac{x}{x_{0}})^n*e^(\frac{-x}{x_{0}}) [/tex] Where A, n, and X0 are constants.

‎[tex]\psi(x) = A(\frac{x}{x_{0}})^ne^(\frac{-x}{x_{0}}) [/tex] Where A, n, and X0 are constants.

[From: ] [author: ] [Date: 11-12-21] [Hit: ]

-Your equation is unreadable, so dont be surprised if you dont get many answers!I dont know if I could solve it anyway as my maths is very rusty. But I would start by finding d²ψ/dx².Simplify the resulting equation then inspect it to look for solutions giving V(infinity) = 0.(Solutions could be simple functions such as V = k/x or V = k/x²).......

Using Schrodinger's equation, find the potential U(x) and energy E such that the given wave function is an eigenfunction (we can assume that at x = infinity there is 0 potential).

Can anyone provide the answer?

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Your equation is unreadable, so don't be surprised if you don't get many answers!

I don't know if I could solve it anyway as my maths is very rusty. But I would start by finding d²ψ/dx². Then substitute this into the time-independent equation:
-[h_bar²/(2m)]d²ψ/dx² + V(x)ψ(x) = Eψ(x)

Simplify the resulting equation then inspect it to look for solutions giving V(infinity) = 0. (Solutions could be simple functions such as V = k/x or V = k/x²).

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keywords: and,are,lrm,psi,tex,Where,frac,constants,‎[tex]\psi(x) = A*(\frac{x}{x_{0}})^n*e^(\frac{-x}{x_{0}}) [/tex] Where A, n, and X0 are constants.

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