Question

The photoelectric work function energy for manganese metal is 3.76 eV (per electron). What wavelength of radiation must be directed at the surface of manganese metal in order to eject electrons electrons having a maximum kinetic energy of 2.00×10 ^–19 Joules/electron.

[Note: The electron volt (eV) is a common energy unit for work function energies. 1 eV = 1.602×10 ^–19 Joules.]

Answer 1

0 D 192 . The given problem will be solved by using the photoelectric equation. Which is given by: hu = 8+ Ek - (i) Where hva energy of radiation d = work function Ek = kinetic energy Number of electrons in manganese atom = 25 0= work function = 25x3.76ey - quer [lev = 1.602x10''5] $- quer - 94% 1.602X10'' I = 1.51% 10"} Eks kinetic energy = 25$ 2.00x po 19 g = 5.0x1085 plugging the values in eqh lil we get hv - 1:51x1017) + S-ox 10189 hç - 1.51 X10"5 + 5.0x1010s [h= planck's constant ] = 6.626x16-39J$ [(= 3.0x 10 %ms") 6.626 X10 A $ * 3x 10°max = 10'7 g 51:51 + 0.5] - .-26 6.626 X3 X 10m = -17 2.01 X 10' .- 26 do 6.626x3x10 m = 9.89 x 10 m - 9.89 nm 2.01 × 1517