Photoelectric Effect: Which Answer is Correct?

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SUMMARY

The discussion centers on calculating the kinetic energy of an electron ejected by a photon of wavelength 250 nm, with the electron exhibiting a de Broglie wavelength of 0.85 nm. Two methods for calculating kinetic energy (KE) are presented: the first uses the equation KE = 0.5mv² and the second employs the de Broglie wavelength formula. The correct kinetic energy calculated using the first method is 3.34 x 10-19 J, while the second method, despite minor errors, also leads to a valid result of 6.87 x 10-4 J when corrected.

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Homework Statement



A photon of wavelength 250 nm ejects an electron from a metal. The ejected electron has a de Broglie wavelength of 0.85 nm.

(a) Calculate the kinetic energy of the electron.

Homework Equations



KE = .5mv^2

The Attempt at a Solution



Which answer is correct?

ANSWER1: E = .5mv^2 and v = h/m[tex]\lambda[/tex]
Therefore, E = .5m(h/[tex]\lambda[/tex])^2

Substituting values (including [tex]\lambda[/tex] = .85*10^(-9)), you get 3.34*10^(-19) J

ANSWER2: wavelength=h/p=h/rad(2mK)
250e-9=6.63e-34/(2*9.11e-31K)
K=6.87e-4 Joules
 
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The first one is correct, I think. But the second method will also give you the correct answer if you do it properly - you just made a couple of little mistakes (one algebraic, one plugging in the wrong number).
 
Ah! Thank you very much!
 

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