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The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

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प्रश्न

The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

संख्यात्मक
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उत्तर

Photoelectric cut-off voltage, V0 = 1.5 V

The maximum kinetic energy of the emitted photoelectrons is given as:

Ke = eV0

Where,

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

∴ Ke = 1.6 × 10−19 × 1.5

= 2.4 × 10−19 J

Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is 2.4 × 10−19 J.

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पाठ 11: Dual Nature of Radiation and Matter - EXERCISES [पृष्ठ २८९]

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एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 11 Dual Nature of Radiation and Matter
EXERCISES | Q 11.3 | पृष्ठ २८९
एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 11 Dual Nature of Radiation and Matter
Exercise | Q 11.3 | पृष्ठ ४०७

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(b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?


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