Question

The minimum wavelength of x-rays produced by an electron beam accelerated through a potential difference of v volt, is equal to
1. \(\frac{e \nu }{hc}\)
2. \(\frac{hc}{e \nu}\)
3. \(\frac{h}{e \nu}\)
4. \(\frac{\nu}{he}\)

Answer 1

Correct Answer - Option 1 : \(\frac{e \nu }{hc}\)

Concept:

• The energy associated with the electron in motion is given as 

\(E = \frac{hc}{λ }\) -- (1)

h is the plank's constant and λ  is de Broglie wavelength, which is the wavelength associated with the matter-wave.

• The energy associated with the electron moving along potential difference v is given as 

E = ev --- (2)

Explanation:

If we equate equation Eq (1) and (2) we get

\(ev = \frac{hc}{λ}\)

\(λ = \frac{hc}{ev}\)

So, the correct option is \(\frac{hc}{e \nu}\)

The de-Broglie wavelength (l) is given by

\(λ= \frac{{h}}{mv}\)

where m is the mass of the particle

v is the velocity of the particle