Final answer:
To find the maximum speed of emitted electrons when light with a wavelength of 390 nm causes photoemission with a stopping potential of 0.888 V, we apply the photoelectric effect equations, converting the wavelength to frequency, then using the stopping potential to determine the kinetic energy, and subsequently calculating the maximum electron velocity.
Step-by-step explanation:
The question asks for the maximum speed of emitted electrons when light with a wavelength of 390 nm shines on a metal surface, causing photoemission, with the stopping potential given as 0.888 V. To solve this, we use the photoelectric equation which relates the kinetic energy (KE) of ejected photoelectrons to the photon energy and the work function (Φ) of the metal: KE = hv - Φ, where h is Planck's constant and v is the frequency of the incident light.
The stopping potential Vs is the potential needed to stop the fastest moving electrons, and it relates to the kinetic energy by eVs = KE, where e is the charge of the electron.
First, we convert the wavelength of light to frequency using c = λv, where c is the speed of light, λ is the wavelength, and v is the frequency. Then, we calculate the kinetic energy (KE) using the stopping potential (eVs = KE), and finally, we find the maximum speed (vmax) of the electrons using the formula: KE = 0.5mvmax^2, where m is the electron mass.