A horizontal cesium plate ( = 1.9 eV ) travels vertically downward with a steady speed v inside a room filled with radiation of wavelength 250 nm and greater. Determine the minimum value of v such that the vertically upward velocity component remains nonpositive for every photoelectron.

Step-by-step solution

The maximum energy of the incident photons corresponds to the minimum wavelength = 250 nm . E = hc = 1240 eV nm 250 nm = 4.96 eV According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons relative to the plate is: K_ max = E - = 4.96 eV - 1.9 eV = 3.06 eV K_ max = 3.06 1.6 10⁻¹⁹ J = 4.896 10⁻¹⁹ J The maximum velocity of the photoelectrons relative to the plate is: u_ max = 2 K_ max m_e = 2 4.896 10⁻¹⁹ 9.1 10⁻³¹ 1.04 10^6 m/s The velocity of a photoelectron relative to t