## Question about 1.33 Part C

$\lambda=\frac{h}{p}$

Annah Khan 1B
Posts: 28
Joined: Fri Jul 22, 2016 3:00 am

### Question about 1.33 Part C

The velocity of an electron that is emitted from a metallic surface by a photon is 3.6 x 10^3 km/s.
(a) What is the wavelength of the ejected electron?
(b) No electrons are emitted from the surface of the metal until the frequency of the radiation reaches 2.50 x 10^16 Hz. How much energy is required to remove the electron from the metal surface?
(c) What is the wavelength of the radiation that caused photoejection of the electron?

I am confused as to how to go about solving part C. What equation is necessary to solve it, and how does it have anything to do with kinetic energy?

Dabin Kang 1B
Posts: 22
Joined: Fri Jun 23, 2017 11:39 am
Been upvoted: 1 time

### Re: Question about 1.33 Part C

Use $E\left ( photon \right )=E\left ( threshold \right )+E_{k}$

Find the energy of the photon:
$E\left ( photon \right )=E\left ( threshold \right )+\frac{1}{2}m_{e}v_{e}^{2}$

$E\left ( photon \right )=\left ( 1.66\times 10^{-17} J\right )+\frac{1}{2}\left ( 9.11\times 10^{-31} kg\right )\left ( 3.6\times 10^{6} m/s\right )^{2}$

$E\left ( photon \right )=2.25\times 10^{-17} J$

Use the energy of the photon to solve for frequency:
$E=h\nu \rightarrow \nu =\frac{E}{h}$

$\nu =\frac{2.25\times 10^{-17} J}{6.626\times 10^{-34} J\cdot s}$

$\nu =3.396\times 10^{16} Hz$

Solve for wavelength:
$c=\lambda \nu \rightarrow \lambda =\frac{c}{\nu }$

$\lambda =\frac{3.00\times 10^{8} m/s}{3.396\times 10^{16} Hz}$

$\lambda =8.83\times 10^{-9} m$