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......\( eV\)
Pick an option to check your answer.
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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