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Structure of Atom question

2019 · 12 Jan · Shift 1 · Q19
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Structure of Atom question

2019 · 12 Jan · Shift 1 · Q19

JEE MainChemistryStructure of AtomMCQ+4 / −1
What is the work function of the metal if the light of wavelength 4000 A∘\mathop A\limits^ \circA∘​ generates photoelectrons of velocity 6 ×\times× 105 ms–1 from it ? (Mass of electron = 9 ×\times× 10–31 kg; Velocity of light = 3 ×\times× 108 ms −-− 1 Plank's constant = 6.626 ×\times× 10–34 Js; Charge of electron = 1.6 ×\times× 10–19 JeV–1)
  1. A
    4.0 eV
  2. B
    0.9 eV
  3. C
    2.1 eV
  4. D
    3.1 eV
View written solutionFree

Correct answer: C

  1. Use Einstein’s photoelectric equation

hν=ϕ+Kmax⁡h\nu = \phi + K_{\max}hν=ϕ+Kmax​

where:

  • ϕ\phiϕ = work function
  • Kmax⁡=12mv2K_{\max} = \dfrac{1}{2}mv^2Kmax​=21​mv2
  • ν=cλ\nu = \dfrac{c}{\lambda}ν=λc​

So,

ϕ=hcλ−12mv2\phi = \frac{hc}{\lambda} - \frac{1}{2}mv^2ϕ=λhc​−21​mv2

  1. Given data

λ=4000 A˚=4000×10−10 m=4×10−7 m\lambda = 4000\,\text{\AA} = 4000 \times 10^{-10}\,\text{m} = 4 \times 10^{-7}\,\text{m}λ=4000A˚=4000×10−10m=4×10−7m

v=6×105 m s−1v = 6 \times 10^5\,\text{m s}^{-1}v=6×105m s−1

h=6.626×10−34 J sh = 6.626 \times 10^{-34}\,\text{J s}h=6.626×10−34J s

c=3×108 m s−1c = 3 \times 10^8\,\text{m s}^{-1}c=3×108m s−1

m=9×10−31 kgm = 9 \times 10^{-31}\,\text{kg}m=9×10−31kg

1 eV=1.6×10−19 J1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}1eV=1.6×10−19J

  1. Calculate energy of incident photon

E=hcλE = \frac{hc}{\lambda}E=λhc​

E=(6.626×10−34)(3×108)4×10−7E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{4 \times 10^{-7}}E=4×10−7(6.626×10−34)(3×108)​

E=19.878×10−264×10−7E = \frac{19.878 \times 10^{-26}}{4 \times 10^{-7}}E=4×10−719.878×10−26​

E=4.9695×10−19 JE = 4.9695 \times 10^{-19}\,\text{J}E=4.9695×10−19J

Convert into eV:

E=4.9695×10−191.6×10−19≈3.106 eVE = \frac{4.9695 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.106\,\text{eV}E=1.6×10−194.9695×10−19​≈3.106eV

  1. Calculate kinetic energy of emitted electron

Kmax⁡=12mv2K_{\max} = \frac{1}{2}mv^2Kmax​=21​mv2

Kmax⁡=12(9×10−31)(6×105)2K_{\max} = \frac{1}{2}(9 \times 10^{-31})(6 \times 10^5)^2Kmax​=21​(9×10−31)(6×105)2

Kmax⁡=12(9×10−31)(36×1010)K_{\max} = \frac{1}{2}(9 \times 10^{-31})(36 \times 10^{10})Kmax​=21​(9×10−31)(36×1010)

Kmax⁡=12(324×10−21)K_{\max} = \frac{1}{2}(324 \times 10^{-21})Kmax​=21​(324×10−21)

Kmax⁡=162×10−21=1.62×10−19 JK_{\max} = 162 \times 10^{-21} = 1.62 \times 10^{-19}\,\text{J}Kmax​=162×10−21=1.62×10−19J

Convert into eV:

Kmax⁡=1.62×10−191.6×10−19≈1.01 eVK_{\max} = \frac{1.62 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.01\,\text{eV}Kmax​=1.6×10−191.62×10−19​≈1.01eV

  1. Find work function

ϕ=E−Kmax⁡\phi = E - K_{\max}ϕ=E−Kmax​

ϕ=3.106−1.01=2.096 eV\phi = 3.106 - 1.01 = 2.096\,\text{eV}ϕ=3.106−1.01=2.096eV

ϕ≈2.1 eV\phi \approx 2.1\,\text{eV}ϕ≈2.1eV

  1. Check options
  • A: 4.0 eV4.0\,\text{eV}4.0eV ❌
  • B: 0.9 eV0.9\,\text{eV}0.9eV ❌
  • C: 2.1 eV2.1\,\text{eV}2.1eV ✅
  • D: 3.1 eV3.1\,\text{eV}3.1eV ❌

Therefore, the correct option is C.

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