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Structure of Atom question

2023 · 12 Apr · Shift 1 · Q16
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Structure of Atom question

2023 · 12 Apr · Shift 1 · Q16

JEE MainChemistryStructure of AtomNumerical+4 / −1

Values of work function (W 0_00​) for a few metals are given below

Metal Li Na K Mg Cu Ag
W 0_00​/eV 2.42 2.3 2.25 3.7 4.8 4.3

The number of metals which will show photoelectric effect when light of wavelength 400 nm400 \mathrm{~nm}400 nm falls on it is ‾\underline{\hspace{2cm}}​

Given: h=6.6×10−34 J s\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}h=6.6×10−34 J s

c=3×108 ms−1c=3 \times 10^{8} \mathrm{~ms}^{-1}c=3×108 ms−1

e=1.6×10−19Ce=1.6 \times 10^{-19} \mathrm{C}e=1.6×10−19C

Numerical answer
View written solutionFree

Correct answer: 3

  1. Condition for photoelectric effect

A metal will show photoelectric effect if the energy of the incident photon is at least equal to its work function:

E=hcλ≥W0E = \frac{hc}{\lambda} \geq W_0E=λhc​≥W0​

  1. Calculate photon energy for λ=400 nm\lambda = 400\,\text{nm}λ=400nm

Convert wavelength into meters:

400 nm=400×10−9 m=4×10−7 m400\,\text{nm} = 400 \times 10^{-9}\,\text{m} = 4 \times 10^{-7}\,\text{m}400nm=400×10−9m=4×10−7m

Now,

E=hcλ=(6.6×10−34)(3×108)4×10−7E = \frac{hc}{\lambda} = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{4 \times 10^{-7}}E=λhc​=4×10−7(6.6×10−34)(3×108)​

E=19.8×10−264×10−7=4.95×10−19 JE = \frac{19.8 \times 10^{-26}}{4 \times 10^{-7}} = 4.95 \times 10^{-19}\,\text{J}E=4×10−719.8×10−26​=4.95×10−19J

Convert this into eV:

E=4.95×10−191.6×10−19 eVE = \frac{4.95 \times 10^{-19}}{1.6 \times 10^{-19}}\,\text{eV}E=1.6×10−194.95×10−19​eV

E≈3.09 eVE \approx 3.09\,\text{eV}E≈3.09eV

  1. Compare with given work functions

Photon energy ≈3.09 eV\approx 3.09\,\text{eV}≈3.09eV

  • Li: 2.42 eV2.42\,\text{eV}2.42eV, so 3.09>2.423.09 > 2.423.09>2.42 ✅
  • Na: 2.30 eV2.30\,\text{eV}2.30eV, so 3.09>2.303.09 > 2.303.09>2.30 ✅
  • K: 2.25 eV2.25\,\text{eV}2.25eV, so 3.09>2.253.09 > 2.253.09>2.25 ✅
  • Mg: 3.70 eV3.70\,\text{eV}3.70eV, so 3.09<3.703.09 < 3.703.09<3.70 ❌
  • Cu: 4.80 eV4.80\,\text{eV}4.80eV, so 3.09<4.803.09 < 4.803.09<4.80 ❌
  • Ag: 4.30 eV4.30\,\text{eV}4.30eV, so 3.09<4.303.09 < 4.303.09<4.30 ❌
  1. Count the metals showing photoelectric effect

The metals are:

Li, Na, K\text{Li, Na, K}Li, Na, K

So the number of metals is

3\boxed{3}3​

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