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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Physics
Chapter
Dual Nature of Matter

BITSAT 2015 · Physics · Dual Nature of Matter

When 0.50 Ã X-rays strike a material, the photoelectrons from the k...

When 0.50A~0.50 Ã X-rays strike a material, the photoelectrons from the k\mathrm{k} shell are observed to move in a circle of radius 23mm23 \mathrm{ mm} in a magnetic field of 2×10−22 \times 10^{-2} tesla acting perpendicularly to the direction of emission of photoelectrons. What is the binding energy of k-shell electrons?

Options

  1. A3.5keV3.5 \mathrm{keV}
  2. B6.2keV6.2 \mathrm{keV}
  3. C2.9keV2.9 \mathrm{keV}
  4. D5.5keV5.5 \mathrm{keV}

Correct answer

B. 6.2keV6.2 \mathrm{keV}

Step-by-step solution

As we know, F=\mathrm{F}=\mathrm{qvB}=\mathrm{m}\mathrmv2R⇒v=qmBR\frac{\mathrmv^{2}}{\mathrm{R}} \Rightarrow \mathrm{v}=\frac{\mathrm{q}}{\mathrm{m}} \mathrm{BR} The kinetic energy of the photoelectron =12mv2=12e2B2R2m=12(1.6×10−19)2(2×10−2)2(23×10−3)2(9.1×10−31)=2.97×10−15J=2.97×10−151.6×10−19=18.36keV\begin{array}{l} =\frac{1}{2} m v^{2}=\frac{1}{2} \frac{e^{2} B^{2} R^{2}}{m}\\ =\frac{1}{2} \frac{\left(1.6 \times 10^{-19}\right)^{2}\left(2 \times 10^{-2}\right)^{2}\left(23 \times 10^{-3}\right)^{2}}{\left(9.1 \times 10^{-31}\right)}\\ =2.97 \times10^{-15} \mathrm{ J}\\ =\frac{2.97 \times 10^{-15}}{1.6 \times 10^{-19}}=18.36 \mathrm{keV} \end{array} Fnergy of the incident n=hcλ=12.40.50=24.8keV\mathrm{n}=\frac{\mathrm{hc}}{\lambda}=\frac{12.4}{0.50}=24.8 \mathrm{keV} Therefore, Binding energy =24.8−18.6=6.2keV=24.8-18.6=6.2 \mathrm{keV}
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