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Apply the relevant quantum physics expressions. (cont.) The stopping potential is AV, 2.11 V so that KE max - eAV e(2.11
Posted: Fri May 06, 2022 1:09 pm
by answerhappygod
- Apply The Relevant Quantum Physics Expressions Cont The Stopping Potential Is Av 2 11 V So That Ke Max Eav E 2 11 1 (19.69 KiB) Viewed 35 times
Apply the relevant quantum physics expressions. (cont.) The stopping potential is AV, 2.11 V so that KE max - eAV e(2.11 V) 2.11 eV. To convert from electron volts (eV) to joules (3), recall that 1 eV 1.60-10-19 J. KE max = e(2.11 V) (1.60 x 10-19 3)(2.11 eV) 3.38 x 10-19 J. The unknown quantity to be determined in part (b) is the maximum photoelectron speed. The photoelectron's maximum kinetic energy is given by KE max W where v is the maximum speed of the electron. What is v, the photoelectron's maximum speed? m/s