question_answer

The de-Broglie wavelength associated with an electron accelerated through a potential difference V is \[\lambda .\] What will be its wavelength when the accelerating potential increased to 4 V?

Answer:

As \[\lambda \propto (1/\sqrt{V}),\] hence on increasing accelerating potential from V to 4V, de-Broglie wavelength decreases from \[\lambda \] to \[\lambda /2.\]