Consider an atomic nucleus to be equivalent to a one dimensional infinite potential well with L=1.4×10-14, a typical nuclear diameter. What would be the ground-state energy of an electron if it were trapped in such a potential well? (Note: Nuclei do not contain electrons.)

Short Answer

Expert verified

1.9 GeV

Step by step solution

01

Given data

The width of the potential well is

L=1.4×10-14m

02

Energy in a potential well

The ground state energy of a particle of mass m in an infinite potential well of width L is

E0=h28meL2n2............1

Here h is the Planck's constant having value

h=6.63×10-34J.s

03

Determining the ground state energy of electron

The mass of the electron is

me=9.11×10-31kg

From equation (I) the ground state energy of electron is

E1=6.63×10-34J.s28×9.11×10-31kg×1.4×10-14m212=3.07×10-10×1J.1J×1kg.m2/s21J.1s2.11kg.11m2=3.07×10-10J

The energy in electron volt is

E1=3.07×10-10×1J×0.624×1019eV1J=1.9×109eV=1.9GeV

The required energy is 1.9GeV.

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Most popular questions from this chapter

Figure 39-9 gives the energy levels for an electron trapped in a finite potential energy well 450 eV deep. If the electron is in the n = 3 state, what is its kinetic energy?

As Fig. 39-8 suggests, the probability density for the region X>L in the finite potential well of Fig. 39-7 drops off exponentially according toψ2(x)=Ce-2kx , where C is a constant. (a) Show that the wave functionψ(x) that may be found from this equation is a solution of Schrödinger’s equation in its one-dimensional form. (b) Find an expression for k for this to be true.

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What is the probability that an electron in the ground state of the hydrogen atom will be found between two spherical shells whose radii are r and r + r, (a) if r = 0.500a and r=0.010aand (b) if r = 1.00a and r=0.01a, where a is the Bohr radius? (Hint: r is small enough to permit the radial probability density to be taken to be constant between r and r+r.)

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