For emission line of atomic hydrogen from n_i=8 to n_f=n, the plot of wave number (v ) against ( 1n^2 ) will be (The Rydberg constant, R_H is in wave number unit)

Question

Accepted Answer

Correct answer: C. (d) As we know that, v=-R_H( 1n_2^2- 1n_2^2)Z^2 (where, Z=1) After putting the values, we get v=-R_H( 1n_2- 18^2) ⇒v= R_H64- R_Hn^2 Comparing to y= mx+c, we get x= 1n^2and m=-R_H (slope)

$For emission line of atomic hydrogen from n_{i} = 8 to n_{f} = n, the plot of wave number (v) against (\frac{1}{n ^{2}}) will be (The Rydberg constant, R_{H} is in wave number unit)$

Solution

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