A) \[1:4\]
B) \[1:2\]
C) \[4:1\]
D) \[2:1\]
Correct Answer: C
Solution :
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
You need to login to perform this action.
You will be redirected in
3 sec
