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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Physics
Chapter
Dual Nature of Matter

BITSAT 2011 · Physics · Dual Nature of Matter

Light of wavelength 180 nm ejects photoelectron from a plate of a...

Light of wavelength 180nm180 \mathrm{nm} ejects photoelectron from a plate of a metal whose work function is 2eV2 \mathrm{eV}. If a uniform magnetic field of 5×10−5T5 \times 10^{-5} \mathrm{ T} is applied parallel to plate, what would be theradius of the path followed by electrons ejected normally from the plate with maximum energy?

Options

  1. A1.239m1.239 \mathrm{ m}
  2. B0.149m0.149 \mathrm{ m}
  3. C3.182m3.182 \mathrm{ m}
  4. D2.33m2.33 \mathrm{ m}

Correct answer

B. 0.149m0.149 \mathrm{ m}

Step-by-step solution

If vmax⁡\mathrm{v}_{\max } is the speed of the fastest electron emitted from the metal surface, then hcλ=W0+12mvmax⁡2(6.63×10−34)×(3×108)(180×10−9) and  =2×(1.6×10−19)+12(9.1×10−31)vmax⁡2∴ and v=1.31×106m/s\begin{aligned} \frac{h c}{\lambda}= W_{0}+\frac{1}{2} m v_{\max }^{2} \frac{\left(6.63 \times 10^{-34}\right) \times\left(3 \times 10^{8}\right)}{\left(180 \times 10^{-9}\right)}\\ \text{\text{ and } } =2 \times\left(1.6 \times10^{-19}\right)+\frac{1}{2}\left(9.1 \times10^{-31}\right) v_{\max }^{2} \\ \therefore \quad \text{ and } v=1.31 \times10^{6} \mathrm{ m}/ \mathrm{s} \end{aligned} The radius of the electron is given by r=mvqB=(9.1×10−31)×(1.31×106)(1.6×10−19)×(5×10−9)=0.149m\frac{m v}{q B}=\frac{\left(9.1 \times 10^{-31}\right) \times\left(1.31 \times 10^{6}\right)}{\left(1.6 \times 10^{-19}\right) \times\left(5 \times 10^{-9}\right)}=0.149 \mathrm{ m}
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