Photo Electric Effect

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Question

Light of wavelength 2475 Å 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 $\frac{e}{m} = 1.7 \times {10^{11}})$

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Solution

Radius of circular path described by a charged particle in a magnetic field is given by $r = \frac{{\sqrt {2mK} }}{{qB}}$ ;
where K = Kinetic energy of electron $\Rightarrow K = \frac{{{q^2}{B^2}{r^2}}}{{2m}} = \left( {\frac{e}{m}} \right)\frac{{e{B^2}{r^2}}}{2}$
$= \frac{1}{2} \times 1.7 \times {10^{11}} \times 1.6 \times {10^{ - 19}} \times {\left( {\frac{1}{{\sqrt {17} }} \times {{10}^{ - 5}}} \right)^2} \times {(1)^2}$
$= 8 \times {10^{ - 20}}J = 0.5\,eV$
By using E = W₀+ K_max

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Photo Electric Effect

Remember concepts with our Masterclasses.

Light of wavelength 2475 Å is incident on barium. Photoelectrons emitted describe a circle of radius 100 cm by a magnetic field of flux density Tesla. Work function of the barium is (Given

Radius of circular path described by a charged particle in a magnetic field is given by ;where K = Kinetic energy of electron By using E = W0 + Kmax

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Light of wavelength 2475 Å 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 $\frac{e}{m} = 1.7 \times {10^{11}})$