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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Atomic Physics

NTA Abhyas 2020 · Physics · Atomic Physics

The velocity of the electrons liberated by electromagnetic...

The velocity of the electrons liberated by electromagnetic radiation of wavelength λ=18.0nm\lambda = 18. 0nm, from stationary He+{He}^{+} ions in the ground state, is

Options

  1. A2. 3 \times {10}^{6}\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm{ms}^{-1}
  2. B1. 3 \times {10}^{6}\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm{ms}^{-1}
  3. C2. 3 \times {10}^{5}\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm{ms}^{-1}
  4. D1. 3 \times {10}^{3}\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm{ms}^{-1}

Correct answer

A. 2. 3 \times {10}^{6}\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm{ms}^{-1}

Step-by-step solution

Step I : Determine binding energy of electron :

As we know E=−13.6Z2n2=−13.6(2)2(1)2(For ground state)<br/><br/> and  and =–54.4eV<br/><br/>∴ and Thebindingenergyofelectronis and E<sub>b</sub>=–E=54.4eV<br/><br/><strong>StepII: and </strong>Determinekineticenergyofelectron.<br/><br/>∴E = \dfrac{- 1 3 \text{.} 6 Z^{2}}{n^{2}} = \dfrac{- 1 3 \text{.} 6 { \left(2\right)}^{2}}{{ \left(1\right)}^{2}} \left(\text{For ground state}\right)<br /><br /> \text{ and } \text{ and } = – 54.4 eV<br /><br />∴ \text{ and } The binding energy of electron is \text{ and } E<sub>b</sub> = – E = 54.4 eV<br /><br /><strong>Step II : \text{ and } </strong>Determine kinetic energy of electron.<br /><br />∴{T}_{e } = hcλ−Eb<br/><br/>=1242nmeV18nm−54.4eV<br/><br/>=(69–54.4)eV=14.6eV<br/><br/>=23.36×10<sup>–19</sup>joule<br/><br/>∴\dfrac{h c}{\lambda} - {E }_{b}<br /><br /> =\dfrac{1 2 4 2 n \text{meV}}{1 8 n m } -5 4 \text{.}4 \text{eV}<br /><br /> = (69 – 54.4) eV = 14.6 eV<br /><br /> = 23.36 × 10<sup>–19</sup> joule<br /><br />∴{T }_{e } = 12mv2<br/><br/>\dfrac{1}{2} {mv}^{2}<br /><br /> v = 2Tem<br/><br/> and  and =2×23.36×10−199.1×10−31m/s=2.3×106m/s\sqrt{\dfrac{2 Te }{m }}<br /><br /> \text{ and } \text{ and } =\sqrt{\dfrac{2 \times 2 3 \text{.} 3 6 \times 1 {0}^{- 1 9}}{9 \text{.} 1 \times 1 {0}^{- 3 1}}} \text{m/s} =2 \text{.}3 \times1 {0}^{6} \text{m/s}
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