In collidge tube the potential difference between cathode and anticathode is 120kV. The maximum energy of emitted X rays will be
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The maximum energy of emitted X rays will be 1.2 x 10^4 ev
Explanation:
Given data:
Potential difference between cathode and anti cathode = 126 KV
To Find:
The maximum energy of emitted X-X−rays will =?
Maximum K.E = 1 /2 mV^2
= 1/2 × eV
= 1/2 x 120 x 10^3
= 1.2 x 10^4 ev
Hence The maximum energy of emitted X rays will be 1.2 x 10^4 ev
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