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

(D) Linear with slope $-\mathrm{R}_{\mathrm{H}}$

(A) Linear with intercept $-\mathrm{R}_{\mathrm{H}}$

(C) Linear with slope $\mathrm{R}_{\mathrm{H}}$

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Atomic Structure

For emission line of atomic hydrogen from

n_{i} = 8

n_{f} = n

, the plot of wave number

(\overline{v})

against

(\frac{1}{n ^{2}})

will be (The Rydberg constant,

R_{H}

is in wave number unit)

Answer: (D)

Write the Rydberg relation with the fixed initial level

n_{i} = 8

and the variable final level

n

\overline{v} = R_{H} (\frac{1}{8 ^{2}} - \frac{1}{n ^{2}}) = - R_{H} \cdot \frac{1}{n ^{2}} + \frac{R _{H}}{64}

. Plotted against

\frac{1}{n ^{2}}

this is a straight line, so the graph is linear with slope

- R_{H}

and intercept

\frac{R _{H}}{64}

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option