JEE MAIN - Chemistry (2019 - 9th January Morning Slot - No. 17)
For emission line of atomic hydrogen from ni = 8 to nf = n, the plot of wave number $$\left( {\overline v } \right)$$ against $$\left( {{1 \over {{n^2}}}} \right)$$ will be (The Rydberg constant, RH is in wave number unit)
Linear with intercept $$-$$RH
Non linear
Linear with slope RH
Linear with slope $$-$$RH
Explanation
We know,
$$\overline v = {1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_f^2}} - {1 \over {n_i^2}}} \right]$$
here ni $$=$$ 8 and nf $$=$$ n
Z $$=$$ 1 for hydrogen
$$ \therefore $$ $$\overline v = {1 \over \lambda } = {R_H}\left[ {{1 \over {{n^2}}} - {1 \over {{8^2}}}} \right]$$
So, graph is $$ \to $$
_9th_January_Morning_Slot_en_17_1.png)
So, graph will be linear with slope RH.
$$\overline v = {1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_f^2}} - {1 \over {n_i^2}}} \right]$$
here ni $$=$$ 8 and nf $$=$$ n
Z $$=$$ 1 for hydrogen
$$ \therefore $$ $$\overline v = {1 \over \lambda } = {R_H}\left[ {{1 \over {{n^2}}} - {1 \over {{8^2}}}} \right]$$
So, graph is $$ \to $$
So, graph will be linear with slope RH.
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