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Question 2

# The work function of caesium metal is 2.14 eV. When light of frequency 6 ×1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the(a) maximum kinetic energy of the emitted electrons,(b) Stopping potential, and(c) maximum speed of the emitted photoelectrons?

Work function of caesium metal, ø0 = 2.14 eV

Frequency of light, v = 6.0 x 1014 Hz

(a) The maximum kinetic energy is given by the photoelectric effect as: K = hv - ø0

Where,

h = Planck’s constant = 6.626 × 10−34 Js

∴ K = 6.626 x 1034 x 6 x 1014 / 1.6 x 10-19 - 2.14 = 2.485 - 2.14 = 0.345 eV

Hence, the maximum kinetic energy of the emitted electrons is 0.345 eV.

( b ) For stopping potential , V0 we can write the equation for kinetic energy as: K = eV0

∴ V0 = K/e = 0.345 x 1.6 x 10-19 / 1.6 x 10-19 = 0.345 V

Hence, the stopping potential of the material is 0.345 V.

(c) Maximum speed of the emitted photoelectrons = v

Hence, the relation for kinetic energy can be written as: K = 1/2 mv2

Where,

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

v2 = 2K/m = 2 x 0.345 x 1.6 x 10-19 / 9.1 x 10-31 = 0.1104 x 1012

∴ v = 3.323 x 105 m/s = 332.3 km/s

Hence, the maximum speed of the emitted photoelectrons is 332.3 km/s.

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The text passage contains a physics problem related to the photoelectric effect, specifically asking for the maximum kinetic energy, stopping potential, and maximum speed of emitted photoelectrons when light of a given frequency is incident on a caesium metal surface. The passage provides detailed calculations and answers for each part of the question.
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Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
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- MULTIPLE_QUESTIONS
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Home Board Paper Solutions Text Book Solutions Articles NCERT Exemplar Solutions
Question 2

# The work function of caesium metal is 2.14 eV. When light of frequency 6 ×1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the(a) maximum kinetic energy of the emitted electrons,(b) Stopping potential, and(c) maximum speed of the emitted photoelectrons?

Work function of caesium metal, ø0 = 2.14 eV

Frequency of light, v = 6.0 x 1014 Hz

(a) The maximum kinetic energy is given by the photoelectric effect as: K = hv - ø0

Where,

h = Planck’s constant = 6.626 × 10−34 Js

∴ K = 6.626 x 1034 x 6 x 1014 / 1.6 x 10-19 - 2.14 = 2.485 - 2.14 = 0.345 eV

Hence, the maximum kinetic energy of the emitted electrons is 0.345 eV.

( b ) For stopping potential , V0 we can write the equation for kinetic energy as: K = eV0

∴ V0 = K/e = 0.345 x 1.6 x 10-19 / 1.6 x 10-19 = 0.345 V

Hence, the stopping potential of the material is 0.345 V.

(c) Maximum speed of the emitted photoelectrons = v

Hence, the relation for kinetic energy can be written as: K = 1/2 mv2

Where,

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

v2 = 2K/m = 2 x 0.345 x 1.6 x 10-19 / 9.1 x 10-31 = 0.1104 x 1012

∴ v = 3.323 x 105 m/s = 332.3 km/s

Hence, the maximum speed of the emitted photoelectrons is 332.3 km/s.

## 3 Comment(s) on this Question

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The text passage contains a physics problem related to the photoelectric effect, specifically asking for the maximum kinetic energy, stopping potential, and maximum speed of emitted photoelectrons when light of a given frequency is incident on a caesium metal surface. The passage provides detailed calculations and answers for each part of the question.

TYPE: SINGLE_QUESTION_WITH_ANSWER