Light of wavelength \(2475\,\text{\AA}\) is incident on barium. Photoelectrons emitted describe a circle of radius \(100\,\text{cm}\) in a magnetic field of flux density \(\frac{1}{\sqrt{17}} \times 10^{-5}\) tesla. The value of work function of barium is _ _ _ _ _ eV. (Given \(e/m = 1.7 \times 10^{11}\))
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Use \(E(\text{eV}) = \frac{12400}{\lambda(\text{\AA})}\) and \(K(\text{eV}) = \frac{1}{2}\frac{e}{m}B^2 r^2\) for quick solving.
Step 1: Understanding the Concept:
This problem combines Einstein's photoelectric equation with the mechanics of a charged particle in a magnetic field. The kinetic energy of the photoelectrons can be found from the radius of their circular path. Step 2: Key Formula or Approach:
1. Einstein's Equation: \(hf = \Phi + K_{max}\).
2. Radius in magnetic field: \(r = \frac{mv}{eB} \implies v = \frac{eBr}{m}\).
3. Kinetic energy: \(K_{max} = \frac{1}{2}mv^{2} = \frac{e}{2} \left( \frac{e}{m} \right) B^{2}r^{2}\). Step 3: Detailed Explanation:
1. Energy of incident photon (\(E\)):
\[ E = \frac{12375}{\lambda (\text{\AA})} = \frac{12375}{2475} = 5\text{ eV} \]
2. Kinetic energy of photoelectrons (\(K_{max}\)):
Radius \(r = 100\text{ cm} = 1\text{ m}\).
Magnetic field \(B = \frac{10^{-5}}{\sqrt{17}}\text{ T}\).
\[ K_{max}\text{ (in Joules)} = \frac{1}{2}m \left( \frac{eBr}{m} \right)^{2} = \frac{1}{2} \left( \frac{e}{m} \right) e B^{2}r^{2} \]
Convert to eV (divide by \(e\)):
\[ K_{max}\text{ (in eV)} = \frac{1}{2} \left( \frac{e}{m} \right) B^{2}r^{2} \]
\[ K_{max} = \frac{1}{2} (1.7 \times 10^{11}) \left( \frac{10^{-5}}{\sqrt{17}} \right)^{2} (1)^{2} \]
\[ K_{max} = \frac{1}{2} \times 1.7 \times 10^{11} \times \frac{10^{-10}}{17} = \frac{0.1 \times 10}{2} = 0.5\text{ eV} \]
3. Work Function (\(\Phi\)):
\[ \Phi = E - K_{max} = 5 - 0.5 = 4.5\text{ eV} \] Step 4: Final Answer:
The work function is 4.5 eV.