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

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

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

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

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