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Structure of Atom question

2022 · 25 Jun · Shift 2 · Q1
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Structure of Atom question

2022 · 25 Jun · Shift 2 · Q1

JEE MainChemistryStructure of AtomMCQ+4 / −1
The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is : [Given : The threshold frequency of platinum is 1.3 ×\times× 1015 s −-− 1 and h = 6.6 ×\times× 10 −-− 34 J s.]
  1. A
    3.21 ×\times× 10 −-− 14 J
  2. B
    6.24 ×\times× 10 −-− 16 J
  3. C
    8.58 ×\times× 10 −-− 19 J
  4. D
    9.76 ×\times× 10 −-− 20 J
View written solutionFree

Correct answer: C

  1. Use Einstein’s photoelectric condition

    The minimum photon energy required to just produce photoelectric emission is the work function: Emin⁡=hν0E_{\min} = h\nu_0Emin​=hν0​ where:

    • h=6.6×10−34 J sh = 6.6 \times 10^{-34}\ \text{J s}h=6.6×10−34 J s
    • ν0=1.3×1015 s−1\nu_0 = 1.3 \times 10^{15}\ \text{s}^{-1}ν0​=1.3×1015 s−1

  2. Substitute the values

    Emin⁡=(6.6×10−34)(1.3×1015)E_{\min} = (6.6 \times 10^{-34})(1.3 \times 10^{15})Emin​=(6.6×10−34)(1.3×1015)

  3. Multiply the numbers

    6.6×1.3=8.586.6 \times 1.3 = 8.586.6×1.3=8.58

  4. Add the powers of 10

    10−34×1015=10−1910^{-34} \times 10^{15} = 10^{-19}10−34×1015=10−19

    Therefore, Emin⁡=8.58×10−19 JE_{\min} = 8.58 \times 10^{-19}\ \text{J}Emin​=8.58×10−19 J

  5. Match with the options

    • A: 3.21×10−14 J3.21 \times 10^{-14}\ \text{J}3.21×10−14 J
    • B: 6.24×10−16 J6.24 \times 10^{-16}\ \text{J}6.24×10−16 J
    • C: 8.58×10−19 J8.58 \times 10^{-19}\ \text{J}8.58×10−19 J
    • D: 9.76×10−20 J9.76 \times 10^{-20}\ \text{J}9.76×10−20 J

    Hence, the correct option is C.

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