First slide
Matter waves (DC Broglie waves)
Question

When a proton is accelerated through a potential difference of 4000 volt, its de Broglie wavelength is λ. What will be the de Broglie wavelength of a deuteron when accelerated through a potential difference of 8000 volt?

Moderate
Solution

p22m=Q.Vp=2mQVλ=hp=h2mQV

λdλp=mpmdQpQd.VpVd=12.1.40008000=12

λd=λ2

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