Question

The wave function of a particle in a box is given by ____________ a) A sin(kx)

b) A cos(kx)

c) Asin(kx) + Bcos(kx)

d) A sin(kx) – B cos(kx)​

Answer 1

Answer:

C) asin (kx) + Bcos(kx)

Answer 2

Answer:

The wave function particle in a 1-D box is given by  Asin(kx)+Bcos(kx).

Therefore, the option (c) is correct.

Explanation:

Consider that a particle of mass 'm' is moving in a one dimensional box of length 'a'. Inside the box potential energy is zero while outside the box it is infinite.

The Schrodinger wave equation for the particle in one dimensional box is:

$\frac{d^2\psi}{dx^2}+ \frac{8\pi ^2m}{h^2}(E-V)\psi =0$

where E is the total energy of the particle.

The potential energy of the particle suddenly increase at x=0 and x=a.

Inside the box V=0, then Schrodinger wave equation becomes:

$\frac{d^2\psi}{dx^2}+ \frac{8\pi ^2m}{h^2}E\psi =0$

ψ measures the position of the particle. ψ must be a continuous function.

$\frac{d^2\psi}{dx^2}+k^2\psi =0$                                                        ....................(1)

where $k^2= \frac{8\pi ^2m}{h^2}E$

For the differential equation (1), the general solution is:

$\psi = A \;sin kx\; +\; B\;cos kx$

where A and B are constants. The value of A and B can be determined by using boundary conditions x=0 and x=a.