Answer to Question #87679 in Quantum Mechanics for Sridhar

Question

A particle is placed inside a one
dimensional box with rigid walls. The first
excited state energy of the particle is
4 eV. The ground state energy is
(A) 2 eV (B) 3 eV
(C) 1 eV (D) 0 eV

Answer 1

(C) 1 eV

$$\frac {W^2_1} {W^2_0}=\frac {2^2} {1^2} (1)$$

Using (1) we got: W₀=1