In a Coolidge tube,the potential difference b | vedclass

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(A) The maximum energy of an $X$-ray photon produced in a Coolidge tube is equal to the kinetic energy gained by an electron accelerated through the p

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$1.2 	imes 10^5 \, eV$

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(A) The maximum energy of an $X$-ray photon produced in a Coolidge tube is equal to the kinetic energy gained by an electron accelerated through the potential difference $V$. Given,potential difference $V = 120 \, kV = 120 	imes 10^3 \, V$. The maximum energy $E_{max}$ is given by the relation: $E_{max} = eV$ Substituting the values: $E_{max} = e 	imes (120 	imes 10^3 \, V) = 120 	imes 10^3 \, eV$ $E_{max} = 1.2 	imes 10^5 \, eV$.

In a Coolidge tube,the potential difference between the cathode and the anode is $120 \, kV$. What will be the maximum energy of the emitted $X$-rays?

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