Question

Scientists have made a light source whose spectral emissive power $\frac{d I}{d \lambda }$ is constant over the visible range. Here $I$ is intensity and $\lambda $ is the wavelength. In other words, $\frac{d I}{d \lambda }$ 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

$\lambda _{t h}=\frac{1240}{2}=620\,nm$ ,
$i=\frac{d N}{d t}e=\frac{e \times A}{h c} \int \lambda dI=$ $\frac{e A k \int\limits _{400}^{620} \lambda d \lambda }{h c}$
Now integrate it with proper limits,
$i=\frac{e A k \int\limits _{400}^{620} \lambda d \lambda }{h c}=\frac{e A k}{h c}\left[\frac{\lambda ^{2}}{2}\right]_{400}^{620}$
$\frac{1 \times 1 \times\left(620^{2}-400^{2}\right) \times k}{1240 \times 2}=\frac{220 \times 1020}{1240 \times 2} \times \frac{0.031}{100}=2.8 \quad A \approx 3 A$