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The potential difference between the cathode and the target electrode in a coolidge tube is 24.75 KV.What is the minimum wavelength of the emitted X rays?
Most Upvoted Answer
The potential difference between the cathode and the target electrode ...
Priyanka want to find wavelength only know...so as per the question... solution is..frequency of emitted x rays is ev=hVmax. therefore Vmax=c÷lambdaev=hc÷lamdaand we want to find lambda right i mean wavelength so;lambda=hc÷ev. (h=6.6 ×10^-34)(c= 3×10^8)(V=24.75 ×103v)(e= 1.6 × 10^-19)so; lambda=6.6×10^-34×3×10^8÷1.6×10^-19×24.75×10^3lambda=19.8×10^-26÷39.6×10^-16so calculate this u get the answer lambda=0.5×10^-10...i get by this manner only...and i think this answer is correct!!..
Community Answer
The potential difference between the cathode and the target electrode ...
It's 0.5 angst. How did u get it suriya??
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The potential difference between the cathode and the target electrode in a coolidge tube is 24.75 KV.What is the minimum wavelength of the emitted X rays?
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