A) 16
B) 32
C) 64
D) 4
Correct Answer: B
Solution :
From Grahams law, \[\frac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\frac{{{M}_{2}}}{{{M}_{1}}}}\] \[\therefore \] \[\frac{\frac{1}{4}{{r}_{{{H}_{2}}}}}{{{r}_{{{H}_{2}}}}}=\sqrt{\frac{{{M}_{{{H}_{2}}}}}{{{M}_{2}}}}\] or \[\frac{1}{4}=\sqrt{\frac{2}{{{M}_{2}}}}\] \[{{M}_{2}}=2\times 16=32\]
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