Question

UV light of wavelength 271 amstrong from a 100 W mercury source eradiates a photocell made of molybdenum metal . If stopping potential is -1.3V estimate work function of the metal. How would the photocell respond to a high intensity (10 power 5 watt per meter square ) red light of wavelength 6328 amstrong produced by helium neon laser ?

Answer 1

Wavelength of ultraviolet light, $\lambda =2271A^{\circ }=2271\times {10}^{-10}m$
Stopping potential of the metal, $V_0=1.3V$
Planck's constant, $h=6.6\times {10}^{-34}J$
Charge on an electron, $e=1.6\times {10}^{-19}C$
Work function of the metal $={\varphi }_0$
Frequency of light = V
We have the photo - energy relation from the photoelectric effect as: $={\varphi }_0$
$=hv-e{V}_0$
$=\frac{hc}{\lambda }-e{V}_0$
$=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{2271\times {10}^{-10}}-1.6\times {10}^{-19}\times 1.3$
$=8.72\times {10}^{-19}-2.08\times {10}^{-19}$
$=6.64\times {10}^{-19}J$
$=\frac{6.64\times {10}^{-19}}{1.6\times {10}^{-19}}=4.15eV$
Let $V_0$ be the threshold frequency of the metal.
$\therefore {\varphi }_0$
$=h{v}_0$
$v_0=\frac{{\varphi }_0}{h}$
$=\frac{6.64\times {10}^{-19}}{6.6\times {10}^{-34}}=1.006\times {10}^{15}Hz$
Wavelength of red light.
${\lambda }_r=6328A^{\circ }$
$=6328\times {10}^{10}m$
$\therefore$ Frequency of red light,
$V_r=\frac{c}{{\lambda }_r}$
$=\frac{3\times {10}^8}{6328\times {10}^{-10}}=4.71\times {10}^{14}Hz$
Since $v_0>v_r$, the photocell will not respond to the red light produced by the laser.