Question:

A student sets up an experiment to investigate the quantum phenomena of photon absorption and emission. They have a sample of atoms in a quantized energy state, where the energy difference between two adjacent energy levels is 2.1 eV. The atoms are in thermal equilibrium with their surroundings at a temperature of 300 K.

a) Calculate the wavelength of the photons emitted when an atom undergoes a transition from the third energy level to the second energy level.

b) Calculate the minimum amount of energy required to remove an electron from an atom in the ground state, if the work function for this material is 3.8 eV.

c) Calculate the maximum velocity with which an electron can be ejected from an atom in the fourth energy level, given that the ionization energy of this level is 6.2 eV.

d) State the energy relationship between the frequency and energy of a photon.

e) A certain metal emits electrons when irradiated with light of a certain frequency. If the intensity of the light is increased, how will this affect the number of ejected electrons?

Answer:

a) The energy difference between two adjacent energy levels can be used to calculate the wavelength of the emitted photon using the equation:

where $\Delta E$ is the energy difference, $h$ is Planck's constant ($6.63 \times 10^{-34} \, \text{J}\cdot\text{s}$), $c$ is the speed of light ($3 \times 10^8 \, \text{m/s}$), and $\lambda$ is the wavelength.

Given $\Delta E = 2.1 \, \text{eV}$, we need to convert it to joules:

$\Delta E = 2.1 \, \text{eV} \times (1.6 \times 10^{-19} \, \text{J/eV}) = 3.36 \times 10^{-19} \, \text{J}$

Substituting the known values into the equation:

Solving for $\lambda$:

Therefore, the wavelength of the photons emitted when an atom undergoes a transition from the third energy level to the second energy level is approximately 6.24 x 10^-7 m.

b) The minimum energy required to remove an electron from an atom is given by the work function, $W$:

Given $W = 3.8 \, \text{eV}$, we need to convert it to joules:

$W = 3.8 \, \text{eV} \times (1.6 \times 10^{-19} \, \text{J/eV}) = 6.08 \times 10^{-19} \, \text{J}$

Therefore, the minimum amount of energy required to remove an electron from an atom in the ground state is approximately $6.08 \times 10^{-19} \, \text{J}$.

c) The maximum velocity of an electron can be determined using the equation:

where $m$ is the mass of the electron and $\text{{ionization energy}}$ is the energy required to remove the electron.

Given $\text{{ionization energy}} = 6.2 \, \text{eV}$, we need to convert it to joules:

$\text{{ionization energy}} = 6.2 \, \text{eV} \times (1.6 \times 10^{-19} \, \text{J/eV}) = 9.92 \times 10^{-19} \, \text{J}$

Rearranging the equation and solving for $v$:

The mass of an electron, $m$, is approximately $9.11 \times 10^{-31} \, \text{kg}$.

Substituting the known values into the equation:

Therefore, the maximum velocity with which an electron can be ejected from an atom in the fourth energy level is approximately $1.14 \times 10^6 \, \text{m/s}$.

d) The energy of a photon is directly proportional to its frequency, $f$, and can be calculated using the equation:

where $E$ is the energy of the photon and $h$ is Planck's constant ($6.63 \times 10^{-34} \, \text{J}\cdot\text{s}$).

Therefore, the energy of a photon is directly proportional to its frequency.

e) The number of ejected electrons is directly proportional to the intensity of the light. If the intensity of the light is increased, the number of ejected electrons will also increase.

Note: Please keep in mind that this is an advanced-level question and should be solved with proficiency in quantum phenomena and related concepts.