(a) A hydrogen atom has its electron in the n = 5 level. The radius of the electron's orbit in the Bohr model is 1.323 nm. Find the de Broglie wavelength of the electron under these circumstances.

(b) What is the momentum, mv, of the electron in its orbit?

To find the de Broglie wavelength of the electron in the given scenario, we can use the formula:

λ = h / p

where λ is the de Broglie wavelength, h is Planck's constant, and p is the momentum of the electron.

To find the momentum of the electron, we can use the equation for the centripetal force in the Bohr model:

F = (mv^2) / r

where F is the electrostatic force between the positively charged nucleus and the negatively charged electron, v is the velocity of the electron, and r is the radius of its orbit.

Since the electron is in the n = 5 energy level, we know that the radius of its orbit, r, is given as 1.323 nm.

The electrostatic force, F, is derived from the equation:

F = k * (Z * e * e) / r^2

where k is Coulomb's constant, Z is the atomic number of the nucleus, and e is the elementary charge.

For hydrogen, Z equals 1, and e is the elementary charge, which is 1.602 x 10^-19 C.

The electrostatic force is balanced by the centripetal force, so we can set F equal to (mv^2) / r and find the velocity, v:

k * (Z * e * e) / r^2 = (mv^2) / r

Rearranging the equation, we find:

v = sqrt((k * Z * e * e) / m)

The mass of an electron, m, is 9.109 x 10^-31 kg.

Now, we can substitute the values into the equation for momentum:

p = m * v

Once we have the value for momentum, we can calculate the de Broglie wavelength using the formula:

λ = h / p

Substituting the known values into the equation will yield the de Broglie wavelength of the electron.

Let me calculate this for you.

(a) To find the de Broglie wavelength of the electron in the given scenario, we can use the equation:

λ = h / p

where λ is the de Broglie wavelength, h is the Planck's constant (6.626 × 10^-34 J·s), and p is the momentum of the electron.

In the Bohr model, the momentum of the electron is given by:

p = mv

where m is the mass of the electron (9.109 × 10^-31 kg) and v is the velocity of the electron.

In the Bohr model, the velocity of the electron can be calculated using the equation:

v = (2.18 × 10^6 m/s) / n

where n is the principal quantum number (in this case, n = 5).

First, let's calculate the velocity of the electron:

v = (2.18 × 10^6 m/s) / 5
v = 4.36 × 10^5 m/s

Now, we can calculate the momentum of the electron:

p = (9.109 × 10^-31 kg)(4.36 × 10^5 m/s)
p = 3.96 × 10^-25 kg·m/s

Finally, we can calculate the de Broglie wavelength of the electron:

λ = (6.626 × 10^-34 J·s) / (3.96 × 10^-25 kg·m/s)
λ ≈ 1.67 × 10^-9 m or 1.67 nm

Therefore, the de Broglie wavelength of the electron in the given scenario is approximately 1.67 nm.

(b) The momentum of the electron in its orbit can be calculated using the equation:

p = mv

where m is the mass of the electron (9.109 × 10^-31 kg) and v is the velocity of the electron.

Using the previously calculated velocity of the electron:

v = 4.36 × 10^5 m/s

Now, we can calculate the momentum of the electron:

p = (9.109 × 10^-31 kg)(4.36 × 10^5 m/s)
p ≈ 3.96 × 10^-25 kg·m/s

Therefore, the momentum (mv) of the electron in its orbit is approximately 3.96 × 10^-25 kg·m/s.