Question

Q35. If width of infinite potential box is reduced by factor 2, energy of particle will be
a Increased by 2 times
b. Decreased by 2 times
c. Increased by 4 times
d. Decreased by 4 times
manath

Answer 1

E = n²h²/(8mL²)

now, L' = L/2

so, E' = n²h²/(8mL'²)

thus, E'/E = L²/(L/2)²

E'/E = 4

thus, E' = 4E

thus, answer is C) Increased by 4 times.

Answer 2

Given:

The Width of the infinite potential box is increased by factor 3  

To find:

Energy pf the particle  

Solution:

1. The relation of the energy is inversely proportional to the square of the width of the potential surface.  

2. The formula used to describe the above relation is,  

=>$E = \frac{n^2h^2}{8mL^2}$,  

=>Only the width factor is changed, and the other values are constants.

=> E = k/L^2,  

3. As the width factor is decreased by a factor of 2, hence the new value of L is,

=> let the changed width factor and energy factor be denoted as L new and E new.

=> L new = (L/2).

4. The value of Energy after changing the width is,

=> E new = k/(L/2)^2,

=> E new = 4k/L = 4(E).

Hence, the energy is increased by 4 times, Option C is the correct answer.