Problem #2
Problem 11.20.b
In an X-ray photoelectron experiment, a photon of wavelength 121 pm eject an electron from the inner shell of an atom and it emerges with a speed of 56.9 Mm/s. Calculate the binding energy of the electron.
The equation for binding energy is .5 m v2 = h ν - Φ
Where m is the mass of the electron, v is the speed of the ejected electron, h is Planck’s constant, ν is the frequency of the photon, and Φ is the binding energy of the electron.
In the problem we are given ν and the wavelength of the photon. Held constant in the problem are m and h. Where m is 9.10938 × 10-31 kg and h is 6.62608 × 10-34.
The speed of the electron is given in Mega meters per second, and by using a conversion factor of 1Mm/(106 m) we can use factor label to convert the speed into meters. Once converted the speed is 5.69 × 107 meters. Therefore λ is 5.69 × 107 meters.
The wavelength is given in picometers, to be able to use c = λ ν to convert wavelength into frequency we must have wavelength in terms of meters. To do this we must use the conversion factor of 1m/(1012 pm), which converts the wavelength into 1.21 × 10-10.
To convert the wavelength of the photon into frequency we must manipulate the equation c = λ ν into c/λ = ν. Using c as the speed of light, which is 2.99792458 × 108, we can do a simply divide c by λ (the wavelength) to get a resulting frequency of 2.47762362 × 1018. So ν is 2.47762362 × 1018.
We know have all the constants but Φ. So if we rearrange the equation by subtracting h ν from both sides, then multiplying the resulting -1 from both sides we end up with.
(.5 m v2 - h f)*(-1) = Φ
Plugging our values into this equation we now have.
((.5 (9.10938 x 10-31)* (5.69 x 107 )2 - (6.62608 x 1034)*(2.47762362 x 1018)) * (-1) = Φ
Which ends up having Φ equal 1.67062243 × 10-16J.
Explanation of Problem #2 with use of Maple
Back to Main Page