Question

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

• A. $1.2\times {10}^{4}eV$
• B. ${10}^{10}eV$
• C. ${10}^{15}eV$
• D. ${10}^{20}eV$

Answer 1

The correct option is A $1.2\times {10}^{4}eV$

Potential difference between cathode and anti cathode $=126KV$

The maximum energy of emitted $X-$rays will $=?$

maximum energy $=\frac{1}{2}{mV}^2$

$=\frac{1}{2}\times eV$

$=\frac{1}{2}\times 120\times {10}^3$

$=1.2\times {10}^{4}eV$