A) 1 : 4
B) 1 : 2
C) 1 : 1
D) 1 : 5
Correct Answer: B
Solution :
| We have \[\frac{1}{2}m{{v}^{2}}_{\max }=E-\phi \] |
| Here case (i) |
| \[\frac{1}{2}m{{v}^{2}}_{1\,\max }=(1-0.5)eV\] |
| \[\frac{1}{2}mv{{_{2}^{2}}_{\,\max }}=(2.5-0.5)eV\] |
| Hence, \[\frac{v_{1}^{2}\max }{v_{2}^{2}\max }=\frac{1}{4}\] |
| \[\frac{{{v}_{1\max }}}{{{v}_{1\max }}}=\frac{1}{2}\] |
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