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anonymous
 one year ago
The minimum frequency for the photoelectric effect of a zinc plate is 9.7 × 10^14 Hz. What will be the kinetic energy of an ejected electron when a photon with a wavelength 200 nm (2.00 × 10^7 m) strikes it?
anonymous
 one year ago
The minimum frequency for the photoelectric effect of a zinc plate is 9.7 × 10^14 Hz. What will be the kinetic energy of an ejected electron when a photon with a wavelength 200 nm (2.00 × 10^7 m) strikes it?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A. 2.00 × 10^19 J B. 3.51 × 10^19 J C. 7.00 × 10^19 J D. 9.94 × 10^19 J

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here we can apply this formula: \[\Large h\nu  h{\nu _0} = KE\] where KE is the requested kinetic energy, \nu_0 = 9.7*10^14 Hz, and \nu is the frequency of the incident photon

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so we have: \[\Large \begin{gathered} KE = \frac{{hc}}{\lambda }  h{\nu _0} = h\left( {\frac{c}{\lambda }  {\nu _0}} \right) = \hfill \\ \hfill \\ = 6.63 \times {10^{  34}}\left( {\frac{{3 \times {{10}^8}}}{{2 \times {{10}^{  7}}}}  9.7 \times {{10}^{14}}} \right) = ...Joules \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\[\large \begin{gathered} KE = \frac{{hc}}{\lambda }  h{\nu _0} = h\left( {\frac{c}{\lambda }  {\nu _0}} \right) = \hfill \\ \hfill \\ = 6.63 \times {10^{  34}}\left( {\frac{{3 \times {{10}^8}}}{{2 \times {{10}^{  7}}}}  9.7 \times {{10}^{14}}} \right) = ...Joules \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok! so we get 3.5139E19? choice B is our solution?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! That's right!
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