Question

Light of intensity ${10}^{-5}{\text{Wm}}^{-2}$ falls on a sodium photo-cell of surface area $2{\text{cm}}^2$. Assuming that the top $5$ layers of sodium absorb the incident energy, estimate the time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about $2\text{eV}$. The atomic area of the sodium atom, $A_e$ is ${10}^{-20}{\text{m}}^2$

• A. ${10}^{-9}\text{s}$
• B. $0.5\text{years}$
• C. $1.6\text{s}$
• D. $1.2\text{years}$

Answer 1

The correct option is B $0.5\text{years}$

Incident power of the light,

$P=I\times A$
$={10}^{-5}\times 2\times {10}^{-4}=2\times {10}^{-9}\text{W}$

Work function of the metal,

${\varphi }_0=2\text{eV}=2\times 1.6\times {10}^{-19}=3.2\times {10}^{-19}\text{J}$

The number of layers of sodium that absorbs the incident energy, $n=5$.

Hence, the number of conduction electrons in n layers is given as :

$n^{\prime }=n\times \left(\frac{A}{A_e}\right)=5\times \left[\frac{(2\times {10}^{-4})}{{10}^{-20}}\right]={10}^{17}$

The incident power is absorbed by all the electrons continuously. The amount of energy absorbed per electron per second is

$E=\frac{P}{n^{\prime }}$

$=\frac{(2\times {10}^{-9})}{{10}^{17}}=2\times {10}^{-26}\text{J/s}$

The time for photoelectric emission
$t=\frac{{\varphi }_0}{E}$
$=\frac{(3.2\times {10}^{-19})}{(2\times {10}^{-26})}=1.6\times {10}^7\text{s}=0.507\text{years}$

Hence, $\left(B\right)$ is the correct answer.