Question

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?

Answer 1

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

Answer 2

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

Answer 3

ok!

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

yes! That's right!

Answer 8

yay! thanks!:)

Answer 9

:)