A) 5893 Å
B) 10683 Å
C) 2791 Å
D) 528 Å
Correct Answer: C
Solution :
Since \[{{W}_{0}}=\frac{hc}{{{\lambda }_{0}}}\]; \[\therefore \ \frac{{{({{W}_{0}})}_{T}}}{{{({{W}_{0}})}_{Na}}}=\frac{{{\lambda }_{Na}}}{{{\lambda }_{T}}}\] or \[{{\lambda }_{T}}=\frac{{{\lambda }_{Na}}\times {{({{W}_{0}})}_{Na}}}{{{({{W}_{0}})}_{T}}}\]\[=\frac{5460\times 2.3}{4.5}=2791\ {\AA}\]
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