Question

Light of wavelength $\lambda =400nm$ falls on an isolated neutral metal ball of radius 1 m and work function 2.1 $eV$. Maximum number of photoelectrons that can be achieved from ball is $n\times {10}^7$, then here n is (take hc = 1240 eV nm) .(Write upto two digits after the decimal point)

Answer 1

$E_{max}=(\frac{1240}{400}-2.1)eV=1eV=1.6\times {10}^{-19}J$
$\Rightarrow V_{max}=1V$
When the voltage developed on the surface of the metal ball equals the stopping potential, the ball stops loosing electrons.

$V=\frac{kQ}{R}\Rightarrow \frac{9\times {10}^9\times n\times 1.6\times {10}^{-19}}{1}=1$

$n=\frac{{10}^{10}}{1.6\times 9}=69.44\times {10}^7$