I have this question from an old exam for review. I tried the question but can't get the right answers.
An electron in an electronically excited hydrogen atom undergoes a transition from a 6d to a 2p orbital, resulting in the emission of a photon. The photon strikes a metal surface where it is absorbed,causing an electron to be ejected having a kinetic energy of 1.32×10-19J.
(a) What is the energy (in J) of the photon emitted by the hydrogen atom?
(b)What is the wavelength (in nm) of the photon emitted by the hydrogen atom?
(c) What is the minimum energy needed to remove an electron from the metal surface?
(d) The wavelength (in nm) of the ejected electron?
See your other post below. You gave up too soon.
To solve this problem, we need to use the principles of energy conservation and the relationships between energy and wavelength in the context of atomic transitions.
(a) Energy of the photon emitted by the hydrogen atom:
The energy change of an electron during a transition between energy levels can be determined using the Rydberg formula:
E = Rh * (1/nf^2 - 1/ni^2)
where E is the energy change, Rh is the Rydberg constant (approximately 2.18 × 10^-18 J), and nf and ni are the final and initial energy levels, respectively.
In this case, the electron transitions from the 6d orbital (ni = 6) to the 2p orbital (nf = 2). Plugging the values into the formula, we have:
E = (2.18 × 10^-18 J) * (1/2^2 - 1/6^2)
E = 1.94 × 10^-18 J
Therefore, the energy of the photon emitted by the hydrogen atom is 1.94 × 10^-18 J.
(b) Wavelength of the photon emitted by the hydrogen atom:
The energy of a photon can also be related to its wavelength using the equation:
E = hc / λ
where E is the energy of the photon, h is Planck's constant (approximately 6.63 × 10^-34 J·s), c is the speed of light (approximately 3.00 × 10^8 m/s), and λ is the wavelength.
We can rearrange this equation to solve for the wavelength:
λ = hc / E
Substituting the known values:
λ = (6.63 × 10^-34 J·s × 3.00 × 10^8 m/s) / (1.94 × 10^-18 J)
λ ≈ 1024 nm
Therefore, the wavelength of the photon emitted by the hydrogen atom is approximately 1024 nm.
(c) Minimum energy needed to remove an electron from the metal surface:
The kinetic energy of the ejected electron can be used to determine the minimum energy needed to remove an electron from the metal surface. According to the conservation of energy, we have:
Energy of the photon = Minimum energy to remove an electron + kinetic energy of the ejected electron
Rearranging the equation, we get:
Minimum energy to remove an electron = Energy of the photon - Kinetic energy of the ejected electron
Minimum energy to remove an electron = 1.94 × 10^-18 J - 1.32 × 10^-19 J
Minimum energy to remove an electron ≈ 1.81 × 10^-18 J
Therefore, the minimum energy needed to remove an electron from the metal surface is approximately 1.81 × 10^-18 J.
(d) Wavelength of the ejected electron:
To determine the wavelength of the ejected electron, we can use the de Broglie equation, which relates the wavelength of a particle to its kinetic energy:
λ = h / (m * v)
where λ is the wavelength, h is Planck's constant, m is the mass of the particle, and v is its velocity.
Since we know the kinetic energy of the ejected electron, we can calculate its velocity using the equation:
KE = (1/2) * m * v^2
Rearranging the equation to solve for v, we have:
v = sqrt(2 * KE / m)
Substituting the given values for the kinetic energy and the mass of an electron:
v = sqrt(2 * (1.32 × 10^-19 J) / (9.11 × 10^-31 kg))
v ≈ 3.08 × 10^6 m/s
Finally, substituting the values for v and Planck's constant into the de Broglie equation:
λ = (6.63 × 10^-34 J·s) / ((9.11 × 10^-31 kg) * (3.08 × 10^6 m/s))
λ ≈ 2.43 × 10^-9 m
Converting the wavelength to nanometers:
λ ≈ 243 nm
Therefore, the wavelength of the ejected electron is approximately 243 nm.