Minimum Wavelength of Electron Accelerated in TV at 30,000 V

AI Thread Summary
An electron accelerated through a potential difference of 30,000 V gains energy calculated as E = qv, resulting in 4.8 x 10^-15 J. To find the minimum wavelength, the equation E = hf can be manipulated into E = hc/λ, indicating that wavelength is inversely proportional to energy. The minimum wavelength corresponds to the maximum energy of the electron, which occurs at the end of the acceleration. The initial kinetic energy is close to zero, increasing as the electron accelerates. Thus, the minimum wavelength is achieved when the electron reaches its maximum energy after being accelerated.
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Homework Statement



In a television, an electron is accelerated through a potential difference of 30,000 V. What is the minimum wavelength produced.


Homework Equations


E = qv
E = hf

The Attempt at a Solution



I figured that we should first find total energy.
E = qv => E = (1 x 10^-16)(30,000) = 4.8 x 10^-15

Then I can manipulate the equation E = hf into E = hc/(lambda) to find the wavelength.
I'm not sure if solving this equation will yield the minimum wavelength or the maximum wavelength. What is the case and why?
 
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the wavelength is inversely proportional to energy as you pointed out, so the minimum wavelength will be at maximum energy
 
That makes sense because frequency will be highest (and wavelength smallest) when energy is maxed.
But, assuming that I solved for maximum energy, what would be minimum energy?
 
hmmm, not too sure... but:

the electron is accelerated from close to rest across 30,000V, so it starts with very low KE, clsoe to zero and its speed increases as it accelerates in teh potential difference upt a maximum at the end of the accerating region
 
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