Question

If you hit the surface of Iron with a photon of energy and find that the ejected electron has a wavelength of .75 nm, what is the wavelength of the incoming photon in nanometers?

Answer 1

Answer:

The wavelength of the incoming photon is 172.8 nm

Explanation:

The wavelength of the incoming photon can be calculated with the photoelectric equation:

$KE = h\frac{c}{\lambda_{p}} - \phi$   (1)

Where:

KE: is the kinetic energy of the electron

h: is Planck's constant = 6.62x10⁻³⁴ J.s  

c: is the speed of light = 3.00x10⁸ m/s

$\lambda_{p}$: is the wavelength of the photon =?  

Φ: is the work function of the surface (Iron) = 4.5 eV        

The kinetic energy of the electron is given by:

$KE = \frac{p^{2}}{2m} = \frac{(\frac{h}{\lambda_{e}})^{2}}{2m}$  (2)

Where:  

p: is the linear momentum = h/λ

m: is the electron's mass = 9.1x10⁻³¹ kg

$\lambda_{e}$: is the wavelength of the electron = 0.75 nm = 0.75x10⁻⁹ m

Hence, the wavelength of the photon is:

$\frac{(\frac{h}{\lambda_{e}})^{2}}{2m} = h\frac{c}{\lambda_{p}} - \phi$

$\lambda_{p} = \frac{hc}{\frac{h^{2}}{2m\lambda_{e}^{2}} + \phi} = \frac{6.62 \cdot 10^{-34} J.s*3.00\cdot 10^{8} m/s}{\frac{(6.62 \cdot 10^{-34} J.s)^{2}}{2*9.1 \cdot 10^{-31} kg*(0.75 \cdot 10^{-9} m)^{2}} + 4.5 eV*\frac{1.602 \cdot 10^{-19} J}{1 eV}} = 1.728 \cdot 10^{-7} m = 172.8 nm$      

Therefore, the wavelength of the incoming photon is 172.8 nm.

I hope it helps you!