Quantum wavefunction (Question 2 (10 Marks)) — UPSC Physics Optional CSE 2024 Paper II
Buy the PYQ solution manualupsc.abhi-physics.com/pyq/upsc-physics-cse-2024-p2-quantum-wavefunction-2
Find the energy of the particle of mass m moving in a potential field V(x) = \frac{2\hbar^2 b^2 x^2}{m} for which the time independent wave function is \psi(x) = \exp(-bx^2). Here b is a constant.
Answers
No answers yet.