question_answer 8)

When the electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy \[1.68\times {{10}^{5}}J\,mo{{l}^{-1}}\]. What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted?

Answer:

Information shadow: Wavelength of incident radiation = 300 nm \[=300\times {{10}^{-9}}m\] Kinetic energy of photoelectrons \[=1.68\times {{10}^{5}}J/mol\] Problem solving strategy: Absorbed = Threshold + Kinetic energy energy per mole energy of photoelectrons of electrons per mole Working it out: Energy of one mole photons = \[Nh\frac{c}{\lambda }\] \[=\frac{6.023\times {{10}^{23}}\times 6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{300\times {{10}^{-9}}}\] \[=3.99\times {{10}^{5}}Jmo{{l}^{-1}}\] Threshold energy \[=3.99\times {{10}^{5}}-1.68\times {{10}^{5}}\] \[=2.31\times {{10}^{5}}J\,mo{{l}^{-1}}\] \[\therefore \]Minimum energy needed to remove electron from the surface of sodium metal \[=2.31\times {{10}^{5}}mo{{l}^{-1}}\] Minimum energy required for one electron \[=\frac{2.31\times {{10}^{5}}}{6.023\times {{10}^{23}}}\] \[=3.84\times {{10}^{-19}}J\] \[\lambda =\frac{hc}{E}\] \[=\frac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{3.84\times {{10}^{-19}}}\] \[=517\times {{10}^{-9}}m=517\] nm(green light)