Question

Scientists have made a light source whose spectral emissive power $\frac{dI}{d\lambda }$ is constant over the visible range. Here $I$ is intensity and $\lambda$ is the wavelength. In other words, $\frac{dI}{d\lambda }$ graph is shown below. This beam of area $1m^2$ is incident on the emitter plate of the photoelectric tube. The collector plate is sufficiently positive so that tube is in saturation mode. Assume that each capable photon liberates $1$ electron. If work function is $2eV,$ what is the current (in $A$ ) in the tube? Round off to the nearest integer.

Answer 1

Answer: 3

${\lambda }_{th}=\frac{1240}{2}=620nm$ ,
$i=\frac{dN}{dt}e=\frac{e\times A}{hc}\int \lambda dI=$ $\frac{eAk\underset{400}{\overset{620}{\int }}\lambda d\lambda }{hc}$
Now integrate it with proper limits,
$i=\frac{eAk\underset{400}{\overset{620}{\int }}\lambda d\lambda }{hc}=\frac{eAk}{hc}{\left[\frac{{\lambda }^2}{2}\right]}_{400}^{620}$
$\frac{1\times 1\times \left(620^2-400^2\right)\times k}{1240\times 2}=\frac{220\times 1020}{1240\times 2}\times \frac{0.031}{100}=2.8A\approx 3A$