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Dual Nature of Radiation question

2022 · 26 Jun · Shift 2 · Q63
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  5. /2022 · 26 Jun · Shift 2 · Q63

Dual Nature of Radiation question

2022 · 26 Jun · Shift 2 · Q63

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
A metal surface is illuminated by a radiation of wavelength 4500 Ao\mathop A\limits^oAo​. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90 ∘^\circ∘ with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately :
  1. A
    1.36 eV
  2. B
    1.69 eV
  3. C
    2.78 eV
  4. D
    2.23 eV
View written solutionFree

Correct answer: A

  1. Use the magnetic field to find the electron speed

Since the electron enters the magnetic field perpendicular to it, it moves in a circle of radius rrr.

For circular motion in a magnetic field: qvB=mv2rqvB = \frac{mv^2}{r}qvB=rmv2​ So, v=qBrmv = \frac{qBr}{m}v=mqBr​

Given:

  • B=2 mT=2×10−3 TB = 2\,\text{mT} = 2 \times 10^{-3}\,\text{T}B=2mT=2×10−3T
  • r=2 mm=2×10−3 mr = 2\,\text{mm} = 2 \times 10^{-3}\,\text{m}r=2mm=2×10−3m
  • q=e=1.6×10−19 Cq = e = 1.6 \times 10^{-19}\,\text{C}q=e=1.6×10−19C
  • m=9.1×10−31 kgm = 9.1 \times 10^{-31}\,\text{kg}m=9.1×10−31kg

Thus, v=(1.6×10−19)(2×10−3)(2×10−3)9.1×10−31v = \frac{(1.6 \times 10^{-19})(2 \times 10^{-3})(2 \times 10^{-3})}{9.1 \times 10^{-31}}v=9.1×10−31(1.6×10−19)(2×10−3)(2×10−3)​ v≈7.03×105 m/sv \approx 7.03 \times 10^5\,\text{m/s}v≈7.03×105m/s

  1. Find the maximum kinetic energy of the photoelectron

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

Kmax⁡=12(9.1×10−31)(7.03×105)2K_{\max} = \frac{1}{2}(9.1 \times 10^{-31})(7.03 \times 10^5)^2Kmax​=21​(9.1×10−31)(7.03×105)2 Kmax⁡≈2.25×10−19 JK_{\max} \approx 2.25 \times 10^{-19}\,\text{J}Kmax​≈2.25×10−19J

Convert into eV: Kmax⁡=2.25×10−191.6×10−19≈1.41 eVK_{\max} = \frac{2.25 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.41\,\text{eV}Kmax​=1.6×10−192.25×10−19​≈1.41eV

  1. Find the energy of incident photon

Wavelength: λ=4500 A˚=4500×10−10 m=4.5×10−7 m\lambda = 4500\,\mathring{A} = 4500 \times 10^{-10}\,\text{m} = 4.5 \times 10^{-7}\,\text{m}λ=4500A˚=4500×10−10m=4.5×10−7m

Photon energy: E=hcλE = \frac{hc}{\lambda}E=λhc​ Using hcλ\frac{hc}{\lambda}λhc​ in eV form: E=1240450 eV≈2.76 eVE = \frac{1240}{450}\,\text{eV} \approx 2.76\,\text{eV}E=4501240​eV≈2.76eV

  1. Apply Einstein's photoelectric equation

E=ϕ+Kmax⁡E = \phi + K_{\max}E=ϕ+Kmax​ So, ϕ=E−Kmax⁡\phi = E - K_{\max}ϕ=E−Kmax​ ϕ=2.76−1.41=1.35 eV\phi = 2.76 - 1.41 = 1.35\,\text{eV}ϕ=2.76−1.41=1.35eV

Therefore, the work function is approximately: 1.36 eV\boxed{1.36\,\text{eV}}1.36eV​

  1. Check options
  • A: 1.36 eV1.36\,\text{eV}1.36eV ✅
  • B: 1.69 eV1.69\,\text{eV}1.69eV
  • C: 2.78 eV2.78\,\text{eV}2.78eV
  • D: 2.23 eV2.23\,\text{eV}2.23eV

Hence, the correct option is A.

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