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Dual Nature of Radiation question

2020 · 9 Jan · Shift 1 · Q64
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Dual Nature of Radiation question

2020 · 9 Jan · Shift 1 · Q64

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
Radiation, with wavelength 6561 Ao\mathop A\limits^oAo​ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of 3 × 10–4 T. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
  1. A
    1.8eV
  2. B
    0.8eV
  3. C
    1.1eV
  4. D
    1.6eV
View written solutionFree

Correct answer: C

  1. Use photoelectric equation

For incident light of wavelength λ=6561 A˚=6561×10−10 m\lambda = 6561\,\text{\AA} = 6561\times 10^{-10}\,\text{m}λ=6561A˚=6561×10−10m,

the photon energy is E=hcλE = \frac{hc}{\lambda}E=λhc​

The maximum kinetic energy of emitted photoelectrons is Kmax⁡=E−ϕK_{\max} = E - \phiKmax​=E−ϕ where ϕ\phiϕ is the work function.

  1. Find maximum kinetic energy from motion in magnetic field

The electron enters a magnetic field B=3×10−4 TB = 3\times 10^{-4}\,\text{T}B=3×10−4T and follows a circular path of largest radius r=10 mm=10−2 mr = 10\,\text{mm} = 10^{-2}\,\text{m}r=10mm=10−2m

For circular motion in a magnetic field, r=mveBr = \frac{mv}{eB}r=eBmv​ so v=eBrmv = \frac{eBr}{m}v=meBr​

Hence, Kmax⁡=12mv2=12m(eBrm)2=e2B2r22mK_{\max} = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{eBr}{m}\right)^2 = \frac{e^2B^2r^2}{2m}Kmax​=21​mv2=21​m(meBr​)2=2me2B2r2​

Substitute values: e=1.6×10−19 C,m=9.1×10−31 kge = 1.6\times 10^{-19}\,\text{C},\quad m = 9.1\times 10^{-31}\,\text{kg}e=1.6×10−19C,m=9.1×10−31kg

Kmax⁡=(1.6×10−19)2(3×10−4)2(10−2)22(9.1×10−31)K_{\max} = \frac{(1.6\times 10^{-19})^2(3\times 10^{-4})^2(10^{-2})^2}{2(9.1\times 10^{-31})}Kmax​=2(9.1×10−31)(1.6×10−19)2(3×10−4)2(10−2)2​

Now, (1.6×10−19)2=2.56×10−38(1.6\times 10^{-19})^2 = 2.56\times 10^{-38}(1.6×10−19)2=2.56×10−38 (3×10−4)2=9×10−8(3\times 10^{-4})^2 = 9\times 10^{-8}(3×10−4)2=9×10−8 (10−2)2=10−4(10^{-2})^2 = 10^{-4}(10−2)2=10−4

So numerator is 2.56×10−38×9×10−8×10−4=23.04×10−50=2.304×10−492.56\times 10^{-38}\times 9\times 10^{-8}\times 10^{-4} = 23.04\times 10^{-50} = 2.304\times 10^{-49}2.56×10−38×9×10−8×10−4=23.04×10−50=2.304×10−49

Denominator: 2×9.1×10−31=1.82×10−302\times 9.1\times 10^{-31} = 1.82\times 10^{-30}2×9.1×10−31=1.82×10−30

Therefore, Kmax⁡=2.304×10−491.82×10−30≈1.266×10−19 JK_{\max} = \frac{2.304\times 10^{-49}}{1.82\times 10^{-30}} \approx 1.266\times 10^{-19}\,\text{J}Kmax​=1.82×10−302.304×10−49​≈1.266×10−19J

Convert to eV: Kmax⁡=1.266×10−191.6×10−19≈0.79 eVK_{\max} = \frac{1.266\times 10^{-19}}{1.6\times 10^{-19}} \approx 0.79\,\text{eV}Kmax​=1.6×10−191.266×10−19​≈0.79eV

Thus, Kmax⁡≈0.8 eVK_{\max} \approx 0.8\,\text{eV}Kmax​≈0.8eV

  1. Find photon energy

Using E=12400λ(A˚) eVE = \frac{12400}{\lambda(\text{\AA})}\,\text{eV}E=λ(A˚)12400​eV

E=124006561≈1.89 eVE = \frac{12400}{6561} \approx 1.89\,\text{eV}E=656112400​≈1.89eV

  1. Calculate work function

ϕ=E−Kmax⁡=1.89−0.79=1.10 eV\phi = E - K_{\max} = 1.89 - 0.79 = 1.10\,\text{eV}ϕ=E−Kmax​=1.89−0.79=1.10eV

So the work function is approximately 1.1 eV\boxed{1.1\,\text{eV}}1.1eV​

  1. Check options
  • A: 1.8 eV1.8\,\text{eV}1.8eV — incorrect
  • B: 0.8 eV0.8\,\text{eV}0.8eV — this is the kinetic energy, not work function
  • C: 1.1 eV1.1\,\text{eV}1.1eV — correct
  • D: 1.6 eV1.6\,\text{eV}1.6eV — incorrect

Therefore, the correct option is C.

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