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  • A light beam of wavelength is incident on a metal having work function of <img alt="1.25 \mathrm{eV}," src="https://entrancecorner.codecogs.com/g

A light beam of wavelength 500 \mathrm{~nm} is incident on a metal having work function of 1.25 \mathrm{eV}, placed in a magnetic field of intensity B. The electrons emitted perpendicular to the magnetic field \mathrm{B}, with maximum kinetic energy are bent into circular arc of radius 30 \mathrm{~cm}. The value of B is _________\times 10^{-7} \mathrm{~T}. Given \mathrm{hc}=20 \times 10^{-26} \mathrm{~J}-\mathrm{m}, mass of electron =9 \times 10^{-31} \mathrm{~kg}
 

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best_answer

\lambda = 500nm= 5\times 10^{-7}m
\phi = 1\cdot 25eV
\frac{hc}{\lambda }= \phi +KE_{max}
\because \frac{1240}{E\left ( eV \right )}= \lambda \left ( nm \right )
\therefore E\left ( eV \right )= \frac{1240}{\lambda \left ( nm \right )}= \frac{1240}{500}= 2\cdot 48\left ( eV \right )
2\cdot 48= 1\cdot 25+KE_{max}
KE_{max}= 1\cdot 23\left ( eV \right )
r= \frac{mv}{q_{B}}= \frac{\sqrt{2mKE}}{q_{B}}
B= \frac{\sqrt{2mKE}}{q {r}}
B= \frac{\sqrt{2\times 9\cdot 1\times 10^{-31}\times 1\cdot 23\times 1\cdot 6\times 10^{-19}}}{1\cdot 6\times 10^{-19}\times 30\times 10^{-2}}
   = \frac{\sqrt{18\cdot 2\times 1\cdot 23\times 1\cdot 6}}{48}\times \frac{10^{-25}}{10^{-21}}
= 0\cdot 12468\times 10^{-4}
= 124\cdot 68\times 10^{-7}J
B= 125\times 10^{-7}J
 

Posted by

vishal kumar

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