When Cesium is illuminated with light of wavelength 546 nm, the maximum speed of the electron emitted is 3.63x10^5 m/s. What is the maximum speed of the emitted electrons when the metal is illuminated by light of wavelength 450 nm?

A hydrogen atom in the n=4 state makes a transition to the ground state, emitting one photon. Calculate the wavelength of the emitted photon and the recoil velocity of the atom.

Please answer......

To determine the maximum speed of an electron emitted from cesium when illuminated with light of a specific wavelength, we can use the principles of the photoelectric effect. This phenomenon occurs when light hits a material and causes the emission of electrons. The key to solving this problem lies in understanding the energy of the incoming photons and how it relates to the kinetic energy of the emitted electrons.

Understanding Photon Energy

The energy of a photon can be calculated using the formula:

E = h * f

where:

• E is the energy of the photon in joules (J),
• h is Planck's constant (approximately 6.626 x 10^-34 J·s), and
• f is the frequency of the light in hertz (Hz).

Relating Wavelength to Frequency

We can also relate the frequency of light to its wavelength using the equation:

c = λ * f

where:

• c is the speed of light (approximately 3.00 x 10⁸ m/s),
• λ is the wavelength in meters (m), and
• f is the frequency in hertz (Hz).

From this equation, we can express frequency as:

f = c / λ

Calculating the Energy of the Photon

Given that the wavelength (λ) is 546 nm, we first convert this to meters:

546 nm = 546 x 10^-9 m

Now, we can find the frequency:

f = c / λ = (3.00 x 10⁸ m/s) / (546 x 10^-9 m) ≈ 5.49 x 10¹⁴ Hz

Next, we calculate the energy of the photon:

E = h * f = (6.626 x 10^-34 J·s) * (5.49 x 10¹⁴ Hz) ≈ 3.64 x 10^-19 J

Determining the Maximum Speed of the Electron

According to the photoelectric effect, the energy of the incoming photon is used to overcome the work function (φ) of the material (cesium in this case) and any excess energy is converted into the kinetic energy (KE) of the emitted electron:

KE = E - φ

The work function for cesium is approximately 2.14 eV, which we need to convert to joules:

φ = 2.14 eV * 1.602 x 10^-19 J/eV ≈ 3.43 x 10^-19 J

Now we can find the kinetic energy of the emitted electron:

KE = E - φ = (3.64 x 10^-19 J) - (3.43 x 10^-19 J) ≈ 0.21 x 10^-19 J

Calculating the Speed

The kinetic energy of an electron can also be expressed in terms of its mass (m) and speed (v):

KE = (1/2) * m * v²

The mass of an electron is approximately 9.11 x 10^-31 kg. Rearranging the formula to solve for speed gives:

v = sqrt((2 * KE) / m)

Substituting the values we have:

v = sqrt((2 * (0.21 x 10^-19 J)) / (9.11 x 10^-31 kg))

Calculating this yields:

v ≈ 1.93 x 10⁶ m/s

Therefore, the maximum speed of the electron emitted from cesium when illuminated with light of wavelength 546 nm is approximately 1.93 million meters per second. This illustrates how light can impart energy to electrons, allowing them to escape from the material, which is a fundamental concept in quantum physics and has numerous applications in technology today.