Question on emission of electrons

AI Thread Summary
Monochromatic light with a wavelength of 4.5 x 10^-7 m can eject electrons from a metal with a maximum kinetic energy (KE) of 3.2 x 10^-19 Joules. To determine if light with a wavelength of 6.8 x 10^-7 m will also cause electron emission, the energy (E) of both wavelengths must be calculated using the equation E = h.c/lambda. The work function (wo) can be derived from the relationship KEmax = E - wo, and it remains constant for the metal. The calculations reveal that while the energy for the longer wavelength is less than the energy for the shorter wavelength, it still exceeds the work function, indicating that emission will occur.
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Homework Statement


Monochromatic light of wavelength 4.5 x 10-7 m will eject electrons from the surface of a
metal with a maximum KE of 3.2 x 10-19 Joules. Will light having a wavelength of 6.8 x 10-7
m also cause emission of electrons from this metal


Homework Equations


E = h.c/lambda
KEmax = E - work function


The Attempt at a Solution


In order for the emission to occur, the E > wo
E = [(6.67x10^-34)(3.0x10^8)]/wavelength
so we can find E for wavelength 4.5x10-7 and for wavelength 6.8x10-7

How do we find wo? (work function)
 
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KEmax = E - work function
In the problem, KEmax and E is given. Find work function using the above equation. Compare this work function with the energy of the new wave length.
 
E = [(6.67x10^-34)(3.0x10^8)]/4.5x10^-7 = 4.446x10-33
since KE = 3.2x10^-19, wo = -3.2x10^-19

Enew = [(6.67x10^-34)(3.0x10^8)]/6.8x10-7 = 2.94 x 10^ -33
since wo is negative, though mag of E is smaller than Wo, E > wo and so there will be emission.

I got these answers but the answer specified in the book is E = 2.9x10^-19 and wo = 1.22x10-19 which I didn't get... any ideas what I am doing wrong?
 
E = [(6.67x10^-34)(3.0x10^8)]/4.5x10^-7 = 4.446x10-33
In this the calculation power is wrong. It should be (-34 + 8 + 7) = ...?
 
Thanks. Since the wo is the same for the metal the one found in 1st case can be used in the 2nd case. Thanks much.
 
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