Question

In collidge tube the potential difference between cathode and anticathode is 120kV. The maximum energy of emitted X rays will be

Answer 1

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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