X-rays of high penetrating power are called hard X-ray. Hard X-rays have energy of the order of `10^5 eV`. The minimum potential difference through which the electrons should be accelerated in an X-ray tube to obtain X-ray of energy `10^5eV` is:

Options

A. `2xx10^(5) V`
B. 50 kV
C. 40 kV
D. `10^(5) V`

Explanation

If electrons are accelerated through a potential difference V, the maximum energy of emitted photon could be
`E_(max)=eV :. 10^(5) eV=eV`
`implies V= 10^(5)V.`