A) \[{{10}^{-6}}\]
B) 10
C) \[{{10}^{-1}}\]
D) \[{{10}^{-10}}\]
Correct Answer: A
Solution :
| [a] \[\frac{{{E}_{A}}}{{{E}_{B}}}=\frac{\frac{hc}{{{\lambda }_{A}}}}{\frac{hc}{{{\lambda }_{N}}}}\] |
| So, \[\frac{{{E}_{A}}}{{{E}_{N}}}=\frac{{{\lambda }_{N}}}{{{\lambda }_{A}}}\] |
| As order of\[{{E}_{A}}=ev\] |
| and order of \[{{E}_{N}}=Mev\] |
| \[\therefore \] \[\frac{{{\lambda }_{N}}}{{{\lambda }_{A}}}=\frac{1ev}{1Mev}=\frac{1}{{{10}^{6}}}={{10}^{-6}}\] |
| So option A is correct Answer |
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