JEE Main · 2022 · Shift-IIeasyATOM-077

The minimum energy that must be possessed by photons to produce the photoelectric effect with platinum metal is:…

Structure of Atom · Class 11 · JEE Main Previous Year Question

Question

The minimum energy that must be possessed by photons to produce the photoelectric effect with platinum metal is:

[threshold frequency =1.3×1015= 1.3\times10^{15} s1^{-1}, h=6.6×1034h=6.6\times10^{-34} Js]

Options
  1. a

    8.58×10198.58\times10^{-19} J

  2. b

    9.76×10209.76\times10^{-20} J

  3. c

    3.21×10143.21\times10^{-14} J

  4. d

    6.24×10166.24\times10^{-16} J

Correct Answera

8.58×10198.58\times10^{-19} J

Detailed Solution

🧠 The Entry Fee (Threshold) The minimum energy (Work Function) is the exact product of hh and the threshold frequency (ν0\nu_0).

🗺️ The Clean Product

  1. Calculation: W0=hν0=(6.6×1034)×(1.3×1015)W_0 = h\nu_0 = (6.6 \times 10^{-34}) \times (1.3 \times 10^{15}).
  2. Result: 8.58×1019 J8.58 \times 10^{-19} \text{ J}.

The 101910^{-19} Range Work functions in Joules are almost always in the 101910^{-19} range. If you get 14-14 or 20-20, you've missed a power of ten.

⚠️ Common Traps Don't use c/λc/\lambda unless wavelength is given. Multiplication is safer here.

Answer: (A)\boxed{\text{Answer: (A)}}

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