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JIPMER 2016
Exam
JIPMER
Year
2016
Subject
Physics
Chapter
Atomic Physics

JIPMER 2016 · Physics · Atomic Physics

An energy of 68.0 eV is required to excite a hydrogen-like atom in...

An energy of 68.0 eV is required to excite a hydrogen-like atom in its second Bohr energy level to third energy level the charge of nucleus is Ze . The wavelength of a radiation required to eject the electron from first orbit to infinity is

Options

  1. A2.2 nm
  2. B2.85 nm
  3. C3.2 nm
  4. D2.5 nm

Correct answer

D. 2.5 nm

Step-by-step solution

Given, n1=2,n2=3n_1=2, n_2=3 and ΔE=68eV\Delta E=68 \mathrm{eV} The difference in energies of two orbits ΔE=13.6Z2[1n12−1n22]\Delta E=13.6 Z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] where, n1<n2n_1 \lt n_2 68=13.6Z2[122−132]=13.6Z2[14−19]=13.6Z2[9−436]=13.6Z2×536\begin{aligned} 68 & =13.6 Z^2\left[\frac{1}{2^2}-\frac{1}{3^2}\right] \\ & =13.6 Z^2\left[\frac{1}{4}-\frac{1}{9}\right] \\ & =13.6 Z^2\left[\frac{9-4}{36}\right] \\ & =13.6 Z^2 \times \frac{5}{36}\end{aligned} ∴Z2=68×3613.6×5=36Z=6\begin{aligned} \therefore \quad Z^2 & =\frac{68 \times 36}{13.6 \times 5}=36 \\ Z & =6\end{aligned} ∵\because Wavelength of photon, 1λ=Z2R[1n12−1n22]\frac{1}{\lambda}=Z^2 R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] n1=1n_1=1 and n2=∞n_2=\infty 1λ=62×R[112−1∞]=36R\frac{1}{\lambda}=6^2 \times R\left[\frac{1}{1^2}-\frac{1}{\infty}\right]=36 R ∴λ=136R=136×1.097×107\therefore \quad \lambda=\frac{1}{36 R}=\frac{1}{36 \times 1.097 \times 10^7} =10−739.5 m=0.025×10−7 m=2.5 nm\begin{aligned} & =\frac{10^{-7}}{39.5} \mathrm{~m} \\ & =0.025 \times 10^{-7} \mathrm{~m} \\ & =2.5 \mathrm{~nm}\end{aligned}
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