Question

A particle of mass ' ( m ) ' is located inside cubic box of size ' ( a ) '. The potential(2) inside the box ( (0 leq x<a ; 0 leq y<a ; 0 leq z<a) )is zero and infinite outside. If the particle is in an eigenstate of energy ( E=frac{14 pi^{2} h^{2}}{2 m a^{2}} ). Show that,[psi(x)=left(frac{2}{a}right)^{frac{1}{2}} sin left(frac{pi x}{a}right) sin left(frac{2 pi y}{a}right) sin left(frac{3 pi z}{a}right)]

          A particle of mass ' ( m ) ' is located inside cubic box of size ' ( a ) '. The potential(2) inside the box ( (0 leq x<a ; 0 leq y<a ; 0 leq z<a) )is zero and infinite outside. If the particle is in an eigenstate of energy ( E=frac{14 pi^{2} h^{2}}{2 m a^{2}} ). Show that,[psi(x)=left(frac{2}{a}right)^{frac{1}{2}} sin left(frac{pi x}{a}right) sin left(frac{2 pi y}{a}right) sin left(frac{3 pi z}{a}right)]

Added by Naman

The Physics of Vibrations and Waves

H. J. Pain 6th Edition

Chapter 9

Instant Answer

Solved by Expert Prabhu Ramji

Step 1

" The potential is zero inside the box and infinite outside, which means the particle is confined within the box. The energy eigenstates for a particle in a 3D box are given by: \[ E_{n_x, n_y, n_z} = \frac{\pi^2 \hbar^2}{2m a^2} (n_x^2 + n_y^2 + n_z^2) \] where Show more…

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A particle of mass ' ( m ) ' is located inside cubic box of size ' ( a ) '. The potential(2) inside the box ( (0 leq x<a ; 0 leq y<a ; 0 leq z<a) )is zero and infinite outside. If the particle is in an eigenstate of energy ( E=frac{14 pi^{2} h^{2}}{2 m a^{2}} ). Show that,[psi(x)=left(frac{2}{a}right)^{frac{1}{2}} sin left(frac{pi x}{a}right) sin left(frac{2 pi y}{a}right) sin left(frac{3 pi z}{a}right)]