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Change in de Broglie wavelength when the accelerating voltage or kinetic energy changes

This type of question, asked 5 times: 2021 Paper 19, 25 Jan 2023, 30 Jan 2023, 15 Apr 2023, 6 Apr 2026 S2. All 5 questions below, oldest first.

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  1. Question 1

    2021 Paper 19

    2
    The de-Broglie wavelength of a particle having kinetic energy EE is λ\lambda. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75%75\% of the initial value?

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  2. Question 2

    25 Jan 2023

    18
    Electron beam used in an electron microscope, when accelerated by a voltage of 20kV20kV, has a deBroglie wavelength of λ0\lambda_{0}. If the voltage is increased to 40kV40kV, then the de-Broglie wavelength associated with the electron beam would be:

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  3. Question 3

    30 Jan 2023

    101
    An electron accelerated through a potential difference V1V_{1} has a de-Broglie wavelength of λ\lambda. When the potential is changed to V2V_{2}, its de-Broglie wavelength increases by 50%50\%. The value of (V1V2)\left( \frac{V_{1}}{V_{2}} \right) is equal to

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  4. Question 4

    15 Apr 2023

    6
    The de Broglie wavelength of an electron having kinetic energy EE is λ\lambda. If the kinetic energy of electron becomes E4\frac{E}{4}, then its de-Broglie wavelength will be:

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  5. Question 5

    6 Apr 2026 S2

    22
    The de Broglie wavelength for an electron accelerated through the potential difference V1V_{1} volt is λ1\lambda_{1}. When the potential difference is changed to V2V_{2} volt, the associated de Broglie wavelength is increased by 50%50\%. If (V1/V2)=(9/α)\left( V_{1}/V_{2} \right) = (9/\alpha), then the value of α\alpha is ____\_\_\_\_ .