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JEE Mains · Physics · STD 12 - Atoms
  • JEE Mains 2014
  • Hard
  • MCQ

The radiation corresponding to \(3 \rightarrow 2\) transition ofhydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of \(3 \times  10^{-4}\) \(T\) If the radius of the largest circular path followed by these electrons is \(10.0\ mm\), the work function of the metal is close to......\( eV\)

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Answer & Solution

Correct Answer

(A) \(1.1\)

Step-by-step Solution

Detailed explanation

Radius of circular path followed by electron is given by,

\(r=\frac{m v}{q B}=\frac{\sqrt{2 m e V}}{e B}=\frac{1}{B} \sqrt{\frac{2 m}{e} V}\)

\(\Rightarrow \quad V=\frac{B^{2} r^{2} e}{2 m}=0.8 \mathrm{V}\)

For transition between \(3\) to \(2\) .

\(E=13.6\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{13.6 \times 5}{36}=1.88 \,eV\)

 Work function \(=1.88 \mathrm{eV}-0.8 \mathrm{eV} \)

\(=1.08 \mathrm{eV} \approx 1.1 \mathrm{eV}\)

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