A) 1/2
B) Zero
C) 3/4
D) 27/32
Correct Answer: C
Solution :
[c] E is minimum when \[{{n}_{1}}=1\And {{n}_{2}}=2\] \[\Rightarrow {{E}_{\min }}=13.6\left( \frac{1}{1}-\frac{1}{4} \right)eV=13.36\times \frac{3}{4}eV\] E is maximum when \[{{n}_{1}}=1\And {{n}_{2}}=\infty \] (the atom is ionized, that is known as ionization energy) \[\Rightarrow {{E}_{\min }}=13.6\left( 1-\frac{1}{\infty }=13.6eV \right)\] \[\therefore \frac{{{E}_{\min }}}{{{E}_{\max }}}=\frac{3}{4}\Rightarrow \frac{hc/{{\lambda }_{\max }}}{hc/{{\lambda }_{\min }}}=\frac{3}{4}\Rightarrow \frac{{{\lambda }_{\max }}}{{{\lambda }_{\min }}}=\frac{3}{4}\]
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