Question

Protons are accelerated from rest by a potential difference 4 kV and strike a metal target. If a proton produces one photon on impact of minimum wavelength ${\lambda }_1$ and similarly an electron accelerated to 4 kV strikes the target and produces a minimum wavelength ${\lambda }_2$ then

• A. ${\lambda }_1={\lambda }_2$
• B. ${\lambda }_1>{\lambda }_2$
• C. ${\lambda }_1<{\lambda }_2$
• D. no such relation can be established

Answer 1

Answer: (A)

${\lambda }_1={\lambda }_2$

If electrons are accelerated to a velocity $v$ by a potential difference V and then allowed to collide with a metal target, the maximum frequency of the X-rays emitted is given by the equation:

$\frac{1}{2}m{v}^2=eV=h\nu$

or, ${\lambda }_{min}=\frac{1240}{V}$.

So this shows that the minimum wavelength is inversely proportional to the accelerating voltage. Here accelerating voltage are same so the minimum wavelength is same.

So, the answer is option (A).