Atomic Structure Result Question 11-1
11. For emission line of atomic hydrogen from $n_i=8$ to $n_f=n$, the plot of wave number ( $v$ ) against $\left(\dfrac{1}{n^2}\right)$ will be (The Rydberg constant, $R_{\mathrm{H}}$ is in wave number unit)
(2019 Main, 9 Jan I)
(a) non linear
(b) linear with slope $-R_{\mathrm{H}}$
(c) linear with slope $R_{\mathrm{H}}$
(d) linear with intercept $-R_{\mathrm{H}}$
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Answer:
Correct Answer: 11. ( c )
Solution:
- According to Rydberg’s formula, wave number $(\bar{\nu})=R_{\mathrm{H}} Z^2\left[\dfrac{1}{n_i^2}-\dfrac{1}{n_f^2}\right]$
Given, $n_i=n, n_f=8 \quad [\because$ it is the case of emission $]$
$\begin{aligned} & \overline{\mathrm{\nu}}=R_{\mathrm{H}} \times(1)^2\left[\dfrac{1}{n^2}-\dfrac{1}{8^2}\right] \\ & \overline{\mathrm{\nu}}=R_{\mathrm{H}}\left[\dfrac{1}{n^2}-\dfrac{1}{64}\right]=\dfrac{R_{\mathrm{H}}}{n^2}-\dfrac{R_{\mathrm{H}}}{64} \end{aligned}$
On comparing with equation of straight line, $y=m x+c$, we get
Slope $=R_{\mathrm{H}}$, intercept $=\dfrac{-R_{\mathrm{H}}}{64}$.
Thus, plot of wave number $(\bar{\nu})$ against $\dfrac{1}{n^2}$ will be linear with slope $\left(+R_{\mathrm{H}}\right)$
BETA
