Question

A single electron atom has nuclear charge $+ Ze$ where $Z$ is atomic number and $e$ is electronic charge. It requires $42.7 eV$ to excite the electron from the second Bohr's orbit to third Bohr's orbit. Find :

(a) the atomic number of element.

(b) the energy required for transition of electron from third to fourth orbit.

(c) the wavelength required to remove electron from first Bohr's orbit to infinity.

(d) the kinetic energy of electron in first Bohr's orbit.

Accepted Answer

(a) $∵ 1 eV = 1.602 \times 10^{- 12} erg$

Also, $ΔE = \frac{hc}{λ} = E_{3} - E_{2} = R_{H} \cdot c \cdot h \cdot Z^{2} [\frac{1}{2^{2}} - \frac{1}{3^{2}}]$

$42.7 \times 1.602 \times 10^{- 12} = 109678 Z^{2} [\frac{1}{2^{2}} - \frac{1}{3^{2}}] \times 3 \times 10^{10} \times 6.626 \times 10^{- 27}$

$∴ Z^{2} = 22.6$

$∴ Z \approx 5$

(b) $ΔE = E_{4} - E_{3} = R_{H} \cdot c \cdot h \cdot Z^{2} [\frac{1}{3^{2}} - \frac{1}{4^{2}}]$

$= 109678 \times 3 \times 10^{10} \times 6.626 \times 10^{- 27} \times 5^{2} \times \frac{7}{16 \times 9}$

$= 26.5 \times 10^{- 12} erg$

(c) $\frac{1}{λ} = R_{H} \cdot Z^{2} [\frac{1}{1^{2}} - \frac{1}{\infty^{2}}]$

$\frac{1}{λ} = 109678 \times 25$

$∴ λ = 3.65 \times 10^{- 7} cm$

(d) $KE = \frac{1}{2} {mu}^{2} = \frac{1}{2} m {(\frac{2 {πZe}^{2}}{nh})}^{2} = \frac{2 π^{2} Z^{2} e^{4} m}{n^{2} h^{2}}$

$= \frac{2 \times {(3.14)}^{2} \times 5^{2} \times {(4.803 \times 10^{- 10})}^{4} \times 9.108 \times 10^{- 28}}{1^{2} \times {(6.625 \times 10^{- 27})}^{2}}$

$= 5.45 \times 10^{- 10} erg$

Embibe Experts Solutions for Chapter: Structure of Atom, Exercise 1: Exercise-1

Embibe Experts Chemistry Solutions for Exercise - Embibe Experts Solutions for Chapter: Structure of Atom, Exercise 1: Exercise-1

Questions from Embibe Experts Solutions for Chapter: Structure of Atom, Exercise 1: Exercise-1 with Hints & Solutions

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