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

Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy?

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प्रश्न

Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (me = 9.11 × 10−31 kg).

संख्यात्मक
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उत्तर

An X-ray probe has greater energy than an electron probe for the same wavelength.

Wavelength of light emitted from the probe, λ = 1 Å = 10−10 m

Mass of an electron, me = 9.11 × 10−31 kg

Planck’s constant, h = 6.6 × 10−34 Js

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

The kinetic energy of the electron is given as:

`"E" = 1/2 "m"_"e""v"^2`

`"m"_"e""v" = sqrt(2"Em"_"e")`

Where,

v = Velocity of the electron

mev = Momentum (p) of the electron

According to the de Broglie principle, the de Broglie wavelength is given as:

`lambda = "h"/"p" = "h"/("m"_"e""v") = "h"/sqrt(2"Em"_"e")`

∴ E = `"h"^2/(2lambda^2"m"_"e")`

`= (6.6 xx 10^(-34))^2/(2xx(10^(-10))^2 xx 9.11 xx 10^(-31))`

= 2.39 × 10−17 J

= `(2.39 xx 10^(-17))/(1.6 xx 10^(-19))`

= 149.375 eV

Energy of a photon `"E'" = "hc"/(lambda"e") "eV"`

= `(6.6 xx 10^(-34) xx 3 xx 10^8) /(10^(-10) xx 1.6 xx 10^(-19))`

= 12.375 × 103 eV

= 12.375 keV

Hence, a photon has greater energy than an electron for the same wavelength.

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पाठ 11: Dual Nature of Radiation and Matter - Exercise [पृष्ठ ४१०]

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एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 11 Dual Nature of Radiation and Matter
Exercise | Q 11.31 | पृष्ठ ४१०

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