Practicing Success
Proton, deutron and a particles are accelerated through the same potential difference. Then the ratio of their de-Broglie wavelength as : |
1 : \(\sqrt { 2}\) : 1 1 : 1 : 1 1 : 1 : \(\sqrt { 2}\) 1 : 1 : 2\(\sqrt { 2}\) |
1 : 1 : 1 |
\(\lambda\) = \(\frac{12.27}{\sqrt{V}}\) A° |