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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Calculate the de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

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प्रश्न

Calculate the de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

संख्यात्मक
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उत्तर

Potential difference, V = 56 V

Planck’s constant, h = 6.6 × 10−34 Js

Mass of an electron, m = 9.1 × 10−31 kg

Charge on an electron, e = 1.6 × 10−19 C

De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:

`lambda = 12.27/sqrt("V")` Å

= `12.27/sqrt56 xx 10^(-10)  "m"`

= 0.1639 nm

Therefore, the de Broglie wavelength of each electron is 0.1639 nm.

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अध्याय 11: Dual Nature of Radiation and Matter - Exercise [पृष्ठ ४०८]

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एनसीईआरटी Physics Part I and II [English] Class 12
अध्याय 11 Dual Nature of Radiation and Matter
Exercise | Q 11.12 (b) | पृष्ठ ४०८

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