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Calculate the maximum kinetic energy and maximum velocity of the photoelectrons emitted when the stopping potential is 81 V for the photoelectric emission experiment. - Physics

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प्रश्न

Calculate the maximum kinetic energy and maximum velocity of the photoelectrons emitted when the stopping potential is 81 V for the photoelectric emission experiment.

संख्यात्मक
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उत्तर

V0 = 81 V

e = 1.6 × 10−19 C

m = 9.1 × 1031 kg

Maximum kinetic energy of electron,

Kmax = `"eV"_0`

= 1.6 × 10−19 × 81

= 129.6 × 1019

= 1.29 × 1017

Kmax = 1.3 × 1017 J

Maximum velocity of photoelectron,

vmax = `sqrt((2"eV"_0)/"m")`

= `sqrt((2 xx 1.6 xx 10^-19 xx 81)/(9.1 xx 10^-31))`

= `sqrt((259.2 xx 10^-19)/(9.1 xx 10^-31))`

= `sqrt(28.48 xx 10^12)`

vmax = 5.3 × 106 ms1

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पाठ 8: Dual Nature of Radiation and Matter - Evaluation [पृष्ठ १३७]

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सामाचीर कलवी Physics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 8 Dual Nature of Radiation and Matter
Evaluation | Q IV. 2. | पृष्ठ १३७

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