A particle is confined to a 1 D box of length 1mm.if the length is changed by 10^(-12)m .the % change in ground state energy is?answer is 2×10^(-4).can u explain this.?

Question

Answer

Explanation:

1 D Box: A one-dimensional box is a hypothetical particle in a box with only one degree of freedom, which can move along a straight line.

Ground State Energy: The energy of the particle in the lowest possible energy state is known as the ground state energy.

Formula: The ground state energy of a 1D box is given by the formula E1= h^2/8mL^2.

Given: Length of 1D box, L = 1mm, Change in length, ΔL = 10^-12m, we need to find the % change in ground state energy.

Solution:

Let's calculate the initial ground state energy, E1.

E1 = h^2/8mL^2

Where h = Planck's constant, m = mass of the particle.

We assume that the particle is an electron.

h = 6.626 x 10^-34 J.s

m = 9.109 x 10^-31 kg

L = 1mm = 10^-3 m

E1 = (6.626 x 10^-34)^2/(8 x 9.109 x 10^-31 x (10^-3)^2) = 1.639 x 10^-18 J

Now, let's calculate the new ground state energy, E2.

We know that the length of the box has changed by ΔL = 10^-12m.

So, the new length of the box, L' = L + ΔL = 1mm + 10^-12m = 1.000000000001mm

L' = 1.000000000001mm = (1 + 10^-12)mm

E2 = h^2/8m(L')^2

E2 = (6.626 x 10^-34)^2/(8 x 9.109 x 10^-31 x (1.000000000001 x 10^-3)^2) = 1.642 x 10^-18 J

The % change in ground state energy is given by the formula:

% Change = |(E2 - E1)/E1| x 100

% Change = |(1.642 x 10^-18 - 1.639 x 10^-18)/1.639 x 10^-18| x 100 = 2 x 10^-4 %

Therefore, the % change in ground state energy is 2 x 10^-4 %.

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