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Chapter 18: Problem 2531

The radio of minimum to maximum wave length in Balmer series is (A) \((1 / 4)\) (B) \((5 / 36)\) (C) \((3 / 4)\) (D) \((5 / 9)\)

Short Answer

Expert verified
The ratio of minimum to maximum wavelength in the Balmer series is \(\frac{5}{9}\).

Step by step solution

01

Identify the equation for the Balmer series

The formula for the Balmer series is given by the Rydberg formula: \[\frac{1}{\lambda} = R_H \left(\frac{1}{2^2} - \frac{1}{n^2}\right)\] where \(\lambda\) is the wavelength, \(R_H\) is the Rydberg constant for hydrogen, and \(n\) is the energy level of the electron. The transitions will occur when the electron falls from higher energy levels to the second energy level where the minimum and maximum wavelengths can be determined.
02

Determine the minimum wavelength

The minimum wavelength corresponds to the highest possible energy transition, which occurs when the electron falls from an infinitely high energy level (\(n \rightarrow \infty\)) to the second energy level (n=2). Substitute \(n \rightarrow \infty\) into the equation to find the minimum wavelength, \(\lambda_{min}\): \[\frac{1}{\lambda_{min}} = R_H \left(\frac{1}{2^2} - \frac{1}{\infty^2}\right)\] \[\frac{1}{\lambda_{min}} = \frac{R_H}{4}\] \[\lambda_{min} = \frac{4}{R_H}\]
03

Determine the maximum wavelength

The maximum wavelength corresponds to the lowest possible energy transition, which occurs when the electron falls from the third energy level (n=3) to the second energy level (n=2). Substitute n=3 into the equation to find the maximum wavelength, \(\lambda_{max}\): \[\frac{1}{\lambda_{max}} = R_H \left(\frac{1}{2^2} - \frac{1}{3^2}\right)\] \[\frac{1}{\lambda_{max}} = R_H \left(\frac{1}{4} - \frac{1}{9}\right)\] \[\frac{1}{\lambda_{max}} = \frac{5R_H}{36}\] \[\lambda_{max} = \frac{36}{5R_H}\]
04

Calculate the ratio of minimum to maximum wavelength

Divide the minimum wavelength by the maximum wavelength to find the ratio: \[\frac{\lambda_{min}}{\lambda_{max}} = \frac{4/R_H}{36/5R_H} = \frac{4}{36}\times\frac{5R_H}{R_H}\] The ratio simplifies to: \[\frac{\lambda_{min}}{\lambda_{max}} = \frac{5}{9}\] Thus, the correct answer is (D) \(\frac{5}{9}\).

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Most popular questions from this chapter

The Probability of survival of a radioactive nucleus for one mean life time is (A) \(1-(1 / \mathrm{e}) \mathrm{S}\) (B) \((1 / \mathrm{e})\) (C) \((2 / \mathrm{e})\) (D) \((3 / \mathrm{e})\)

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which of the following atom has the lowest ionization potential? (A) \({ }^{14}{ }_{7} \mathrm{~N}\) (B) \({ }^{40}{ }_{18} \mathrm{Ar}\) (C) \({ }^{133} 55 \mathrm{Cs}\) (D) \({ }^{16}{ }_{8} \mathrm{O}\)

Match column I and II and chose correct Answer form the given below. (a) Nuclear fusion (p) converts some matter into energy (b) Nuclear fission (q) generally Possible for nuclei with low atomic number (c) \(\beta\) decay (r) generally Possible for nuclei with high atomic number (d) Exothermic nuclear (s) Essentially Proceeds by weak reaction nuclear force(c) (A) $\mathrm{a} \rightarrow \mathrm{p}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{s}, \mathrm{d} \rightarrow \mathrm{q}$ (B) $\mathrm{a} \rightarrow \mathrm{q}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{p}, \mathrm{d} \rightarrow \mathrm{s}$ (C) $\mathrm{a} \rightarrow \mathrm{q}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{s}, \mathrm{d} \rightarrow \mathrm{p}$ (D) $\mathrm{a} \rightarrow \mathrm{r}, \mathrm{b} \rightarrow \mathrm{q}, \mathrm{c} \rightarrow \mathrm{p}, \mathrm{d} \rightarrow \mathrm{s}$
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