First slide

Photo Electric Effect

Question

Light of wavelength 2475 A is incident on barium. Photoelectrons emitted describe a circle of radius 100 cm by a magnetic field of flux density $\frac{1}{\sqrt{17}} \times 10^{- 5}$ Tesla. Work function of the barium is (Given $\frac{e}{m} = 1 .7 \times 10^{11}$ )

Difficult

A

1.8 eV
B

2.1 eV
C

12.5 eV
D

None of these

Solution

Radius of circular path described by a charged particle in a magnetic field is given by $r = \frac{\sqrt{2 mK}}{qB}$ where K is Kinetic energy of electron
$\begin{array}{l} \Rightarrow K = \frac{q^{2} B^{2} r^{2}}{2 m} = (\frac{e}{m}) \frac{{eB}^{2} r^{2}}{2} = \frac{1}{2} \times 1 .7 \times 10^{11} \times 1 .6 \times 10^{- 19} \times {(\frac{1}{\sqrt{17}} \times 10^{- 5})}^{2} \times {(1)}^{2} \\ = 8 \times 10^{- 20} J = 0 .5 eV \\ By using E = W_{o} + K_{\max} \\ \Rightarrow W_{o} = E - K_{\max} (\frac{12375}{2475}) eV - 0 .5 eV = 4 .5 eV \end{array}$

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