Question

If light of wavelength 560 nm strikes such a photocathode, what will be the maximum kinetic energy, in ev , of the emitted electrons?

Answer 1

The maximum kinetic energy
`(\(K_{\text{max}}\))` of emitted electrons when light of wavelength 560 nm strikes the photocathode is approximately 2.21 eV.

Step-by-step explanation:

When light strikes a photocathode, the energy of the incident photons
`(\(E_{\text{photon}}\))` can be determined using the relationship
`\(E_{\text{photon}} = (hc)/(\lambda)\), where \(h\)` is Planck's constant
`(\(6.626 * 10^(-34) \, \text{J} \cdot \text{s}\)), \(c\)` is the speed of light
`(\(3.00 * 10^8 \, \text{m/s}\)), and \(\lambda\)` is the wavelength of light. Substituting the given values, we find
`\(E_{\text{photon}} \approx 3.54 * 10^(-19) \, \text{J}\).`

The work function
`(\(\phi\))` of the photocathode represents the minimum energy required to liberate an electron from the material. The relationship between
`\(K_{\text{max}}\), \(E_{\text{photon}}\), and \(\phi\) is given by \(K_{\text{max}} = E_{\text{photon}} - \phi\). Assuming a typical work function for metals, \(\phi \approx 4.5 \, \text{eV}\)`, we can now calculate
`\(K_{\text{max}}\). Substituting the values, we get \(K_{\text{max}} \approx 2.21 \, \text{eV}\).`

In conclusion, the maximum kinetic energy of the emitted electrons is determined by the difference between the energy of the incident photons and the work function of the photocathode. The energy of the incident photons is calculated using the wavelength of light, and by subtracting the work function, we obtain the maximum kinetic energy of the emitted electrons. In this case, it is approximately 2.21 eV.

Answer 2

The maximum kinetic energy of the ejected electrons can be determined using the equation KE = hf - BE, where KE is the maximum kinetic energy, hf is the energy of the incident photons, and BE is the binding energy or work function of the photocathode material. By plugging in the values provided, the maximum kinetic energy of the emitted electrons can be calculated to be 2.22 eV.

Step-by-step explanation:

The maximum kinetic energy (KE) of the ejected electrons can be determined using the equation:

KE = hf - BE

Where:

• KE is the maximum kinetic energy of the electrons
• hf is the energy of the incident photons, which can be calculated using the equation E = hc/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the light
• BE is the binding energy, also known as the work function, of the photocathode material

In this case, the wavelength of the light is 560 nm. Plugging this value into the equation E = hc/λ gives:

E = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s)/(560 x 10^-9 m) = 3.55 x 10^-19 J

Converting the energy from joules to electron volts (eV) using the conversion factor 1 eV = 1.602 x 10^-19 J, we get:

E = (3.55 x 10^-19 J)/(1.602 x 10^-19 J/eV) = 2.22 eV

Therefore, the maximum kinetic energy of the emitted electrons is 2.22 eV.