Answer to Question #102847 in Electricity and Magnetism for Kishor

Question

emmitor and collector plates are saprated by 10cm are connected through an ammeter without any cell. work function of emmitor is 2.39ev and light on at a wavelength is 400 find the minimum value of megnetic field which current registered by ameter is zero

Answer 1

$\frac{hc}{\lambda}=A+\frac{mv^2}{2} \to v=\sqrt{\frac{2( \frac{hc}{ \lambda } - A)}{m}}$

$v=\sqrt{\frac{2( \frac{hc}{ \lambda } - A)}{m}}=\sqrt{\frac{2( \frac{6.62\cdot 10^{-34}\cdot 3\cdot 10^8}{400\cdot 10^{-9}} - 2.39\cdot 1.6\cdot 10^{-19})}{9.1\cdot 10^{-31}}}=501755 m/s$

$evB=m\frac{v^2}{R} \to B=\frac{mv}{eR}$

$B=\frac{mv}{eR}=\frac{9.1\cdot 10^{-31}\cdot 501755}{1.6\cdot 10^{-19}\cdot 0.1}=2.85\cdot 10^{-5} T$