Mathematically equation (7-22) is the same differential equation as we had for a particle in a box-the function and its second derivative are proportional. But(ϕ)for m1= 0is a constant and is allowed, whereas such a constant wave function is not allowed for a particle in a box. What physics accounts for this difference?

Short Answer

Expert verified

The wave function m1= 0, which has no dependence and satisfies the condition without forcing anything to be zero everywhere.

Step by step solution

01

Significance of quantum numbers

The quantum number is used to describe the trajectory as well as the movement of the electron in an atom. According to the Pauli Exclusion principal, no electron in an atom can have same set of the quantum numbers.

02

Explanation of physics accounts for the difference

In case of the hydrogen atom and other particles in the spherical well, no barrier is encountered by varying the azimuthal angle over the complete range of values from 0 to 2. The wave function is not required to be zero that is encountered and the requirement instead is only that wave function at any angle must return to the same value at an angle that is greater. The wave function for m1 = 0, which has no dependence, already satisfies the condition without forcing anything to be zero everywhere.

Therefore, the wave function m1 = 0 , which has no dependence, and satisfies the condition without forcing anything to be zero everywhere.

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Most popular questions from this chapter

Exercise 80 discusses the idea of reduced mass. When two objects move under the influence of their mutual force alone, we can treat the relative motion as a one particle system of mass μ=m1v2/(m1+m2). Among other things, this allows us to account for the fact that the nucleus in a hydrogen like atom isn’t perfectly stationary, but in fact also orbits the centre of mass. Suppose that due to Coulomb attraction, an object of mass m2and charge -eorbits an object of mass m1 and charge +Ze . By appropriate substitution into formulas given in the chapter, show that (a) the allowed energies are Z2μmE1n2, where is the hydrogen ground state, and (b) the “Bohr Radius” for this system is ma0 ,where a0is the hydrogen Bohr radius.

Explicitly verify that the simple function Rr=Aebrcan be made to satisfy radial equation (7-31), and in so doing, demonstrate what its angular momentum and energy must be.

For the more circular orbits, =n-1and

P(r)r2ne-2r/na0

a) Show that the coefficient that normalizes this probability is

localid="1660047077408" (2na0)2n+11(2n)!

b) Show that the expectation value of the radius is given by

r¯=n(n+12)a0

and the uncertainty by

Δr=na0n2+14

c) What happens to the ratioΔr/r¯in the limit of large n? Is this large-n limit what would be expected classically?

Can the transition 2s1s in the hydrogen atom occur by electric dipole radiation? The lifetime of the 2 s is known to be unusual. Is it unusually short or long?

Using the functions given in Table 7.4, verify that for the more circular electron orbit in hydrogen (i.e.,l=n-1), the radial probability is of the form

P(r)r2ne-2r/nao

Show that the most probable radius is given by

rmostprobable=n2ao

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