Question

A deutron is accelerated through a potential of 500 volts. Find the potential through which a singly ionised helium ion is to be accelerated for the same De Broglie wavelength.​

Answer 1

Answer:very goood

Explanation:

Dbbhajjan

Answer 2

Answer:

The de Broglie wavelength is given by

λ = h

m v .

The rest energy of an electron is about 0.5 MeV and the electrons here have kinetic energies of 15 keV, so the

non-relativistic approximation is OK - we can use the rest mass for m. So we need to find the speed. That can be

found from the kinetic energy which can, in turn, be found from the potential difference V and the change in

potential energy, eV.

So the kinetic energy is K = l

2

m v2 and v = 2K

m

p = mv = m 2K

m = 2mK = 2meV since K = eV

λ = h

2meV

= 6.63 × 1 0-34 J.s

2 × 9.11 × 1 0-31 kg × 1.602 × 1 0-19 C × 15 × 1 03 V

= 1.0 × 10-11

m = 0.010 nm

This is much shorter than visible light wavelengths (about 500 nm). Wave effects would be noticed only on a

scale much smaller than that at which we notice interference and diffraction effects for light. Wave effects would not

be noticed in a TV tube.

1.2 Interference and diffraction effects become

Explanation: