KCET2016PhysicsDual Nature of Matter
The de Broglie wavelength of an electron accelerated to a potential of ( 400 ~V ) is approximately
Options
- A( 0.03 ~nm )
- B( 0.04 ~nm )
- C( 0.12 ~nm )
- D( 0.06 ~nm )
Correct answer
D. ( 0.06 ~nm )
Step-by-step solution
Given, electron is accelerated to a potential of ( 400 ~V ), then de Broglie wavelength is related to potential as [ array l = 1.227 ~nm V = 1.227 ~nm 400 = 1.227 20 ~nm =0.06 ~nm array ] Therefore, de Broglie wavelength is ( 0.06 ~nm ).