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Structure of Atom question

2019 · 9 Jan · Shift 1 · Q15

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JEE Main
Chemistry
Structure of Atom
2019 · 9 Jan · Shift 1 · Q15

Structure of Atom question

2019 · 9 Jan · Shift 1 · Q15

JEE MainChemistryStructure of AtomMCQ+4 / −1

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)

A
Linear with intercept $-$ RH
B
Non linear
C
Linear with slope RH
D
Linear with slope $-$ RH

View written solutionFree

Correct answer: C

Use the Rydberg formula for emission

For hydrogen, the wave number of an emitted line is

\overline{\nu} = R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)

Here, $n_i = 8$ and $n_f = n$ . So,

\overline{\nu} = R_H\left(\frac{1}{n^2}-\frac{1}{8^2}\right)

\overline{\nu} = R_H\left(\frac{1}{n^2}-\frac{1}{64}\right)

Compare with equation of a straight line

Let

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

Then,

\overline{\nu} = R_H x - \frac{R_H}{64}

This is of the form

y = mx + c

with

Slope: $m = R_H$
Intercept: $c = -\frac{R_H}{64}$

Interpret the graph

So the plot of $\overline{\nu}$ versus $\left(\frac{1}{n^2}\right)$ is:

linear
with slope $R_H$

Check options

A: Linear with intercept $-R_H$ ❌

Intercept is actually $-\frac{R_H}{64}$ , not $-R_H$ .
B: Non linear ❌

The equation is linear.
C: Linear with slope $R_H$ ✅
D: Linear with slope $-R_H$ ❌

Slope is positive.

Final Answer

\boxed{\text{C}}

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