Please show work and explain. Answer is already given. Thank You. A certain meta
ID: 1304021 • Letter: P
Question
Please show work and explain. Answer is already given. Thank You.
A certain metal in the photoelectric effect experiment has a work function of 2.3 eV. For a given experiment, the maximum kinetic energy of the ejected electrons is 1.6 eV. What is the wavelength of the incident photons? 218 nm 318 nm 418 nm 518 nm 618 nm What is the DeBroglie wavelength of the ejected electrons? 0.57 nm 0.67 nm 0.77 nm 0.87 nm 0.97 nm What is the maximum wavelength of photons that would eject electrons from this metal? 339 nm 439 nm 539 nm 639 nm 739 nm If theintensity of light was increased, how would the above answer change? it would increase would decrease it would stay the sameExplanation / Answer
work function W =2.3 eV
Kmax = 1.6 eV
17)
from the einstien relation
E = W + Kmax
E = 3.9 eV
hc/L = 3.9x1.6x10^-19
wavelenght L = 6.625x10^-34*3x10^8/(3.9x1.6x10^-19 ) =317.9 nm = 318 nm
b) 318 nm
18)
debroglie wavelenght of the electron L =h/sqrt(2mK)
L =6.625x10^-34/sqrt(2*9.1x10^-31*1.6*1.6*10^-19) = 0.97 x 10^-9 = 0.97 nm
19)
minimum energy needed to eject the electrons is called as work function
corresponding to maximum wavelenght
L =hc/W = 6.625x10^-34*3x10^8/(2.3*1.6x10^-19) = 539 x10^-9 m = 539 nm
c) 539 nm
photo electric effect does not depend on intensity of incident light , it only depends on incident energy
c)
it would stay the same