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

1.66e-9

soln

To find the de Broglie wavelength of the electron in the n=5 level, we can use the formula:

λ = h / p

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

Step 1: Find the momentum (mv) of the electron in its orbit.
In the Bohr model, the electron's momentum is given by:

mv = (nh) / (2πr)

Where n is the principal quantum number, h is Planck's constant, and r is the radius of the electron's orbit.

Given: n = 5 and r = 1.323 nm.

Step 2: Calculate the momentum using the equation:

mv = (5 * h) / (2π * 1.323 nm)

Let's plug in the values and calculate:

mv = (5 * 6.626 x 10^-34 Js) / (2π * 1.323 x 10^-9 m)

Step 3: Simplify the expression:

mv ≈ (5 * 6.626 x 10^-34 Js) / (2 * 3.1416 * 1.323 x 10^-9 m)

mv ≈ 2.513 x 10^-24 kg * m/s

Now that we have the momentum of the electron (mv) as 2.513 x 10^-24 kg * m/s, we can move on to calculating its de Broglie wavelength.

Step 4: Calculate the de Broglie wavelength using the equation:

λ = h / (mv)

Let's plug in the values and calculate:

λ = (6.626 x 10^-34 Js) / (2.513 x 10^-24 kg * m/s)

λ ≈ 2.63 x 10^-10 m

Therefore, the de Broglie wavelength of the electron in the n=5 level is approximately 2.63 x 10^-10 meters. The momentum of the electron is approximately 2.513 x 10^-24 kg * m/s.

To find the de Broglie wavelength of the electron, we can use the equation:

λ = h / p

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

To find the momentum, we can use the equation for the centripetal force acting on the electron:

F = (m*v^2) / r

where F is the centripetal force, m is the mass of the electron (m = 9.11 × 10^-31 kg), v is the velocity of the electron, and r is the radius of the electron's orbit.

Now, let's calculate the de Broglie wavelength and momentum:

First, we need to find the velocity of the electron. In the Bohr model, the electron's velocity can be expressed as:

v = (Ze^2) / (4πε₀nh)

where Z is the atomic number (for hydrogen, Z = 1), e is the elementary charge (e = 1.602 × 10^-19 C), ε₀ is the vacuum permittivity (ε₀ = 8.854 × 10^-12 C^2/(Nm^2)), n is the principal quantum number (n = 5), and h is Planck's constant.

Now, plug in the values:

v = (1 * (1.602 × 10^-19 C)^2) / (4π * (8.854 × 10^-12 C^2/(Nm^2)) * 5 * (6.626 × 10^-34 J⋅s))

Simplifying this expression gives:

v ≈ 2.18 × 10^6 m/s

Now we can calculate the momentum:

p = m * v

p = (9.11 × 10^-31 kg) * (2.18 × 10^6 m/s)

p ≈ 1.98 × 10^-24 kg⋅m/s

Finally, we can calculate the de Broglie wavelength:

λ = h / p

λ = (6.626 × 10^-34 J⋅s) / (1.98 × 10^-24 kg⋅m/s)

λ ≈ 3.34 × 10^-10 m or 334 pm (picometers)

So, the de Broglie wavelength of the electron in this case is approximately 334 picometers (pm), and the momentum of the electron in its orbit is approximately 1.98 × 10^-24 kg⋅m/s.