Question

Some atomic processes can result in the emission of an electron from the atom. Electrons emitted in this way can have discrete values of kinetic energy, which depend on the atomic energy levels. The probability that the electron has a particular value depends on interacions within the atom. Suppose a particular atomic process results in three values of the electron kinetic energy: Ψ1 has Ee = 10 eV, with probability 0.2. Ψ2 has Ee = 16 eV, with probability 0.3. Ψ3 has Ee = 34 eV, with probability 0.5. The electron's wave function can be written as a quantum superposition: Ψ = aΨ1 + bΨ2 + cΨ3 What are the magnitudes of a, b, and c?

Answer 1

Answer:

|a| = 0.447

|b| = 0.547

|c| = 0.707

Step-by-step explanation:

From available data

Ψ1 = Ee = 10eV probability = 0.2

Ψ2 = Ee = 16eV, probability = 0.3

Ψ3 = Ee = 34eV, probability = 0.5

The magnitude of a

= a² = 0.2

= a = √0.2

|a| = 0.447

The magnitude of b

b² = 0.3

b = √0.3

|b| = 0.547

The magnitude of c

c² = 0.5

c = √0.5

|c| = 0.707

So in conclusion we have magnitude of a = 0.447

b = 0.547

c = 0.707

Answer 2

The magnitude of a , b and c are $0.447, 0.547$ and $0.707$ respectively.

Electron kinetic energy :

The electron's wave function is given as,

          $\psi=a\psi _{1}+b \psi _{2}+c\psi _ {3}$

three values of the electron kinetic energy is given as,

$\psi _{1}$ has E = 10 eV, with probability 0.2

$\psi _{2}$ has E = 16 eV, with probability 0.3.

$\psi _{3}$ has E= 34 eV, with probability 0.5.

The value of a, b and c shown below,

    $a^{2}=0.2\\ \\a=\sqrt{0.2}=0.447\\ \\b=\sqrt{0.3}=0.547\\ \\c=\sqrt{0.5}=0.707$

Learn more about the Kinetic energy here:

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