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Thread: Calculate the threshold wavelength of the metal

  1. #1 Calculate the threshold wavelength of the metal 
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    Caesium metal of work function is illuminated with radiations of wavelength . Calculate thei)threshold wavelength of the metal;(ii)maxium energy of liberated elecrons.Since threshold frequency and threshold wavelenght are not the same, am confused of the formula to use.What is the best formula to use? Should I use or should use


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  3. #2  
    Bullshit Intolerant PhDemon's Avatar
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    For the first part calculate the wavelength corresponding to an energy of 2 eV using the second formula (E=hc/lambda), the work function is the minimum energy to cause electron emission, so the result of this calculation is the threshold wavelength. For the second part use the same formula to calculate the energy of the illuminating radiation, the maximum energy of the liberated electrons is the difference between the energy of a photon of the illuminating radiation and the work function.


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  4. #3  
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    For the first part calculate the wavelength corresponding to an energy of 2 eV using the second formula (E=hc/lambda)
    =
    For the second part use the same formula to calculate the energy of the illuminating radiation
    You mean I should still use
    the maximum energy of the liberated electrons is the difference between the energy of a photon of the illuminating radiation and the work function.
    Could the maximum energy be 2.0-2.0 = 0?
    Last edited by chikis; August 14th, 2014 at 03:25 AM.
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  5. #4  
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    For the first part you need to convert 2 eV into joules before putting the numbers in the equation but otherwise the method is correct. For the second part, E=hc/lambda = hc/5e-7, why have you used the answer to part one rather than the radiation wavelength to calculate the energy of the illuminating radiation? If you do these parts again correctly you will see the difference is not zero. Remember if the energy of the incident radiation is less than or equal to the work function no photoelectrons are emitted.
    Last edited by PhDemon; August 14th, 2014 at 11:10 AM. Reason: typo
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  6. #5  
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    ignorance is fixable with data-- stupid is not
    take your pick
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  7. #6  
    Bullshit Intolerant PhDemon's Avatar
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    Who is this^ aimed at? Do you think my advice is incorrect?
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  8. #7  
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    Quote Originally Posted by PhDemon View Post
    For the first part you need to convert 2 eV into joules before putting the numbers in the equation but otherwise the method is correct.
    If the method is correct, I need not to worry. I just got to leave it like that since the method is correct.
    For the second part, E=hc/lambda = hc/5e-7
    From I can clearly see that So how come is By the way, what is 5e?
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  9. #8  
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    Quote Originally Posted by fizzlooney View Post
    ignorance is fixable with data-- stupid is nottake your pick
    My friend, you have not answered the question, who is that aimed at?
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  10. #9  
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    I wouldn't leave it, a correct method/equation with incorrect numbers will not give you the right answer... The 5e-7 was my lazy shorthand for 5 x 10-7 which is the wavelength you were given in the question.
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  11. #10  
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    Quote Originally Posted by PhDemon View Post
    I wouldn't leave it, a correct method/equation with incorrect numbers will not give you the right answer...
    So what are the right numbers?
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  12. #11  
    Bullshit Intolerant PhDemon's Avatar
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    Read my posts again, you need to convert the energy in eV into joules if you want to calcuate a threshold wavelength in metres, FFS look up the conversion factor, or do you expect me to spoon feed you the answer?
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  13. #12  
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    Quote Originally Posted by PhDemon View Post
    Read my posts again, you need to convert the energy in eV into joules
    How do I carry out the conversion?
    or do you expect me to spoon feed you the answer?
    No, not at all, I need only guidance and direction and not the answer.
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  14. #13  
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    Let me google that for you ev to joules - Google Search I don't mind helping but try and at least do some work yourself. I'm trying to guide you to the right answer but you seem to be unable or unwilling to follow the guidance...
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    threshholdwavelenthmaximum energy E = hf-hfo
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  16. #15  
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    The first part is now correct, the second is not. I've told you how to do it. Have another go. Remember the energy of the electron is the energy of the incident radiation (lambda = 5e-7 m) minus the work function.
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  17. #16  
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    Quote Originally Posted by PhDemon View Post
    Remember the energy of the electron is the energy of the incident radiation (lambda = 5e-7 m) minus the work function.
    Are you saying the maximum energy which is given as hf-hfo translates as
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  18. #17  
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    No, the units of the two terms in your latest effort are different, how can that possibly be right? I've told you how to do it at least twice, if you still can't work it out I don't know how to explain it in any simpler terms (and as it looks like homework I'm not doing it for you), I give up on you...
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  19. #18  
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    Quote Originally Posted by PhDemon View Post
    I give up on you...
    Pease don't give up on me yet. My main problem now is hf. How do I get the hf? I know that E=hf-hfo.l
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  20. #19  
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    Last chance, what is the energy (in joules) of a photon with a wavelength of 5e-7 m? [use E=hc/lambda which is the same as E=hf as f=c/lambda]. What is the work function in joules? [using the conversion factor I had to google for you, this is equal to the hf0 term in your last post]. The answer is the difference between these two numbers. I have explained this a number of times and feel as though I am wasting my time on someone unable to understand or unwilling to think and follow my advice.
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