Question

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 \(e/m = 1.7 \times 10^{11}\))

Hint:

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.

Answer 1

Correct Answer: 4.5

Concept:
•Photoelectric equation: \[ E = \phi + K_{\text{max}} \]
•Motion in magnetic field: \[ r = \frac{mv}{eB} \Rightarrow v = \frac{eBr}{m} \]
•Kinetic energy: \[ K = \frac{1}{2}mv^2 = \frac{e^2 B^2 r^2}{2m} \]

Step 1: Given data \[ \lambda = 2475\,\text{\AA}, \quad r = 1\,\text{m} \] \[ B = \frac{1}{\sqrt{17}} \times 10^{-5}\,\text{T}, \quad \frac{e}{m} = 1.7 \times 10^{11} \]

Step 2: Kinetic energy (in eV) \[ K(\text{eV}) = \frac{1}{2} \cdot \frac{e}{m} \cdot B^2 r^2 \] \[ B^2 = \frac{1}{17} \times 10^{-10} \] \[ K = \frac{1}{2} \times 1.7 \times 10^{11} \times \frac{1}{17} \times 10^{-10} \] \[ = \frac{1}{2} \times 0.1 \times 10 = 0.5\,\text{eV} \]

Step 3: Photon energy \[ E = \frac{12400}{\lambda(\text{\AA})} = \frac{12400}{2475} \approx 5\,\text{eV} \]

Step 4: Work function \[ \phi = E - K = 5 - 0.5 = 4.5\,\text{eV} \] Final: 4.5 eV