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Dual Nature of Radiation question

2021 · 24 Feb · Shift 2 · Q47
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  5. /2021 · 24 Feb · Shift 2 · Q47

Dual Nature of Radiation question

2021 · 24 Feb · Shift 2 · Q47

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be :
  1. A
    10 −-− 2 nm
  2. B
    10 −-− 1 nm
  3. C
    10 −-− 3 nm
  4. D
    10 −-− 4 nm
View written solutionFree

Correct answer: C

  1. Use Duane–Hunt law for minimum wavelength

    In an X-ray tube, the maximum photon energy is obtained when the entire kinetic energy of the electron is converted into a single photon: eV=hν=hcλmin⁡eV = h\nu = \frac{hc}{\lambda_{\min}}eV=hν=λmin​hc​

    Hence, λmin⁡=hceV\lambda_{\min} = \frac{hc}{eV}λmin​=eVhc​

  2. Substitute the given voltage

    Given: V=1.24×106 VV = 1.24\times 10^6\ \text{V}V=1.24×106 V

    Using the standard value: hce=1240 eV⋅nm\frac{hc}{e} = 1240\ \text{eV·nm}ehc​=1240 eV⋅nm

    So, λmin⁡=1240 eV⋅nm1.24×106 eV\lambda_{\min} = \frac{1240\ \text{eV·nm}}{1.24\times 10^6\ \text{eV}}λmin​=1.24×106 eV1240 eV⋅nm​

  3. Calculate

    λmin⁡=12401.24×106 nm\lambda_{\min} = \frac{1240}{1.24\times 10^6}\ \text{nm}λmin​=1.24×1061240​ nm

    λmin⁡=10−3 nm\lambda_{\min} = 10^{-3}\ \text{nm}λmin​=10−3 nm

  4. Match with the options

    λmin⁡=10−3 nm\lambda_{\min} = 10^{-3}\ \text{nm}λmin​=10−3 nm

    Therefore, the correct option is: C: 10−3 nm10^{-3}\,\text{nm}10−3nm

  5. Comparison with stored answer

    Stored correct answer: C

    Derived answer: C

    They match.

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