A) \[{{E}_{2}}-{{E}_{1}}=\frac{hc}{\lambda }=3.16\times {{10}^{-19}}J\]
B) \[\Rightarrow \]
C) \[\frac{{{N}_{2}}}{{{N}_{1}}}=\exp \left( \frac{-3.16\times {{10}^{-19}}J}{(1.38\times {{10}^{-23}}1/k).(300k)} \right)\]
D) None of these
Correct Answer: A
Solution :
We have,\[\frac{x}{2}+\frac{y}{3}=1\] \[\frac{x}{2}+\frac{y}{1}=1\]\[({{a}^{2}},-{{b}^{2}})\] \[{{x}^{2}}+9<{{(x+3)}^{2}}<8x+25,\]\[\frac{x+y}{x-y}=\frac{5}{2},\]\[\frac{x}{y}\]\[\frac{3}{8}\] \[\frac{8}{3}\]\[\frac{5}{3}\]\[\frac{3}{5}\]\[6\frac{2}{3}%\]
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