A) \[1:4\]
B) \[1:2\]
C) \[4:1\]
D) \[2:1\]
Correct Answer: C
Solution :
[c] Given de Broglie Wavelength |
\[{{\lambda }_{p}}={{\lambda }_{\alpha }}\] |
So, \[\frac{h}{{{m}_{p}}\times {{v}_{p}}}=\frac{h}{{{m}_{\alpha }}\times {{v}_{\alpha }}}\] |
\[\frac{{{v}_{p}}}{{{v}_{\alpha }}}=\frac{{{m}_{\alpha }}}{{{m}_{p}}}=\frac{4{{m}_{p}}}{{{m}_{p}}}\] |
Because Mass of \[\alpha -particle\] is 4 times mass of proton |
So 4:1 |
Option C is correct answer |
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