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

[zoner7]

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?

 

[lanedance]

the wavelength is inversely proportional to energy as you pointed out, so the minimum wavelength will be at maximum energy

 

[zoner7]

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?

 

[lanedance]

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