A platinum surface having work function 5.63 eV is illuminated by a monochromatic source of 1.6×10^15 Hz. What will be the minimum wavelength associated with the ejected electron.

Kmax = 1 eV; minimum de Broglie wavelength 1.23×10^-9 m (1.23 nm)

PhysicsCBSE Class 12 · Sample paper, session 2024-25 · 2 marks · Very short answer

CBSE Class 12 Physics sample paper 2024-25: question 17

From the official CBSE sample question paper for 2024-25, section B, question 17, for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q17 · 2 marks · Very short answerDual Nature of Radiation and Matter

A platinum surface having work function 5.63 eV is illuminated by a monochromatic source of $1.6 \times 1 0^{15}$ Hz. What will be the minimum wavelength associated with the ejected electron.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

Open one hint at a time and try again before the next.

Hint 1

First find how much energy each photon carries and how much is left after paying the work function.

Hint 2

The wavelength 'associated with the electron' is its de Broglie wavelength, which depends on its momentum.

Hint 3

Use $p = 2 m K$ with the maximum kinetic energy to get the smallest wavelength.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Energy of the incident photon: $E = h ν = 6.63 \times 1 0^{- 34} \times 1.6 \times 1 0^{15} = 1.06 \times 1 0^{- 18} J = \frac{1.06 \times 1 0 ^{- 18}}{1.6 \times 1 0 ^{- 19}} = 6.63 eV$
½ mark
By Einstein's photoelectric equation, the maximum kinetic energy of the ejected electrons is $K_{ma x} = h ν - ϕ_{0} = 6.63 - 5.63 = 1.0 eV = 1.6 \times 1 0^{- 19} J$
½ mark
The electron with the largest kinetic energy has the largest momentum, so it has the minimum de Broglie wavelength: $λ_{min} = \frac{h}{p _{ma x}} = \frac{h}{2 m K _{ma x}}$
½ mark
$λ_{min} = \frac{6.63 \times 1 0 ^{- 34}}{2 \times 9.1 \times 1 0 ^{- 31} \times 1.6 \times 1 0 ^{- 19}} = \frac{6.63 \times 1 0 ^{- 34}}{5.40 \times 1 0 ^{- 25}} \approx 1.23 \times 1 0^{- 9} m = 1.23 nm$
½ mark

Answer: $K_{ma x} = 1$ eV; minimum de Broglie wavelength $\approx 1.23 \times 1 0^{- 9}$ m (1.23 nm)

Where the marks are

Calculates photon energy $h ν \approx 6.63$ eV (½ mark)
Finds $K_{ma x} = h ν - ϕ_{0} = 1$ eV (½ mark)
Uses $λ = h / 2 m K$ for the electron (½ mark)
Correct value $\approx 1.23 \times 1 0^{- 9}$ m (½ mark)

Revise the chapter

Chapter: Dual Nature of Radiation and Matter (notes, formulas, where marks are lost and every past question)
Topic: de Broglie Wavelength

Practise this chapterBoard-style questions from Dual Nature of Radiation and Matter, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2024-25: every question, by section.

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CBSE Class 12 Physics sample paper 2024-25: question 17

Hints

Step-by-step solution

Revise the chapter

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