Question
For emission line of atomic hydrogen from $${n_i} = 8$$ to $${n_f} = n,$$ the plot of wave number $$\left( {\bar v} \right)$$ against $$\left( {\frac{1}{{{n^2}}}} \right)$$ will be (The Rydberg constant, $${R_H}$$ is in wave number unit)
A.
Linear with intercept $$ - {R_H}$$
B.
Non linear
C.
Linear with slope $${R_H}$$
D.
Linear with slope $$ - {R_H}$$
Answer :
Linear with slope $$ - {R_H}$$
Solution :
$$\eqalign{
& {\text{As we know, }} \cr
& \bar v = - {R_H}\left( {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right)\,{Z^2}\,\left( {{\text{where}},\,Z = 1} \right) \cr
& {\text{After putting the values, we get }} \cr
& \bar v = - {R_H}\left( {\frac{1}{{{n^2}}} - \frac{1}{{{8^2}}}} \right) \cr
& \therefore \,\,\bar v = \frac{{{R_H}}}{{64}} - \frac{{{R_H}}}{{{n^2}}} \cr
& {\text{Comparing to y = }}mx{\text{ + }}c{\text{, we get}} \cr
& x{\text{ = }}\,{\text{and}}\,m{\text{ = }}\, - {{\text{R}}_H}\left( {{\text{slope}}} \right) \cr} $$