JEE MAIN - Chemistry (2019 - 9th April Morning Slot - No. 19)
For any given series of spectral lines of atomic
hydrogen, let $$\Delta \mathop v\limits^\_ = $$ $$\Delta {\overline v _{\max }} - \Delta {\overline v _{\min }}$$ be the difference
in maximum and minimum frequencies in
cm–1. The ratio Lyman Balmer $${{\Delta {{\overline v }_{Lyman}}} \over {\Delta {{\overline v }_{Balmer}}}}$$ is :
9 : 4
4 : 1
27 : 5
5 : 4
Explanation
We know,
$$\overline v = R{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$
$$ \therefore $$ For Lyman series,
$${\Delta {{\overline v }_{Lyman}}}$$ = $${\Delta {{\overline v }_{\max }} - \Delta {{\overline v }_{\min }}}$$
= $$\left[ {{1 \over 1} - {1 \over \infty }} \right] - \left[ {{1 \over 1} - {1 \over 4}} \right]$$
= $${{1 \over 4}}$$
$$ \therefore $$ For Balmer series,
$${\Delta {{\overline v }_{Balmer}}}$$ = $${\Delta {{\overline v }_{\max }} - \Delta {{\overline v }_{\min }}}$$
= $$\left[ {{1 \over 4} - {1 \over \infty }} \right] - \left[ {{1 \over 4} - {1 \over 9}} \right]$$
= $${{1 \over 9}}$$
$$ \therefore $$ $${{\Delta {{\overline v }_{Lyman}}} \over {\Delta {{\overline v }_{Balmer}}}}$$ = $${{{1 \over 4}} \over {{1 \over 9}}}$$ = $${{9 \over 4}}$$
$$\overline v = R{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$
$$ \therefore $$ For Lyman series,
$${\Delta {{\overline v }_{Lyman}}}$$ = $${\Delta {{\overline v }_{\max }} - \Delta {{\overline v }_{\min }}}$$
= $$\left[ {{1 \over 1} - {1 \over \infty }} \right] - \left[ {{1 \over 1} - {1 \over 4}} \right]$$
= $${{1 \over 4}}$$
$$ \therefore $$ For Balmer series,
$${\Delta {{\overline v }_{Balmer}}}$$ = $${\Delta {{\overline v }_{\max }} - \Delta {{\overline v }_{\min }}}$$
= $$\left[ {{1 \over 4} - {1 \over \infty }} \right] - \left[ {{1 \over 4} - {1 \over 9}} \right]$$
= $${{1 \over 9}}$$
$$ \therefore $$ $${{\Delta {{\overline v }_{Lyman}}} \over {\Delta {{\overline v }_{Balmer}}}}$$ = $${{{1 \over 4}} \over {{1 \over 9}}}$$ = $${{9 \over 4}}$$
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