## 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?

1. anonymous

A. 2.00 × 10^-19 J B. 3.51 × 10^-19 J C. 7.00 × 10^-19 J D. 9.94 × 10^-19 J

2. Michele_Laino

here 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

3. anonymous

ok!

4. Michele_Laino

so we have: $\Large \begin{gathered} KE = \frac{{hc}}{\lambda } - h{\nu _0} = h\left( {\frac{c}{\lambda } - {\nu _0}} \right) = \hfill \\ \hfill \\ = 6.63 \times {10^{ - 34}}\left( {\frac{{3 \times {{10}^8}}}{{2 \times {{10}^{ - 7}}}} - 9.7 \times {{10}^{14}}} \right) = ...Joules \hfill \\ \end{gathered}$

5. Michele_Laino

$\large \begin{gathered} KE = \frac{{hc}}{\lambda } - h{\nu _0} = h\left( {\frac{c}{\lambda } - {\nu _0}} \right) = \hfill \\ \hfill \\ = 6.63 \times {10^{ - 34}}\left( {\frac{{3 \times {{10}^8}}}{{2 \times {{10}^{ - 7}}}} - 9.7 \times {{10}^{14}}} \right) = ...Joules \hfill \\ \end{gathered}$

6. anonymous

ok! so we get 3.5139E-19? choice B is our solution?

7. Michele_Laino

yes! That's right!

8. anonymous

yay! thanks!:)

9. Michele_Laino

:)