Question

Ultraviolet light of wavelength $2271$ from a $100W$ mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is $-1.3V$, estimate the work function of the metal. How would the photo-cell respond to a high intensity (${10}^{5}W{m}^2$) red light of wavelength $6328$ produced by a He-Ne laser?

Answer 1

Given that,

$\lambda =2271\dot{A}$

$V_0=1.3V$

$so,$

$W=h\nu -e{V}_0$

$=\frac{hc}{\lambda }-e{V}_0$

$=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{2271\times {10}^{-10}}-1.6\times {10}^{-19}\times 1.3$

$=\frac{6.64\times {10}^{-19}}{1.6\times {10}^{-19}}$

$=4.15eV$

Let ${\nu }_0$is the threshold wavelength of metal. So,

${\nu }_0=\frac{W}{h}$

$=\frac{6.64\times {10}^{-19}}{6.6\times {10}^{-34}}$

$=1.006\times {10}^{15}Hz$

Wavelength of red light is

${\lambda }_r=6328\dot{A}=6328\times {10}^{-10}m$

$\therefore {\nu }_r=\frac{c}{{\lambda }_r}$

${\nu }_r=\frac{3\times {10}^8}{6328\times {10}^{-10}}$

$=4.74\times {10}^{14}Hz$

Since, ${\nu }_0\prec {\nu }_r$, so photocell would respond.