Question

A beam of light has three wavelengths $4144\mathring{A}$, $4972\mathring{A}$ and $6216\mathring{A}$ with a total intensity of $3.6\times {10}^{-3}{Wm}^{-2}$ equally distributed among the two wavelengths. The beam fall normally on area of $1{cm}^2$ of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photo electrons liberated in 2s is approximately

• A. $6\times {10}^{11}$
• B. $9\times {10}^{11}$
• C. $11\times {10}^{11}$
• D. $15\times {10}^{11}$

Answer 1

The correct option is C $11\times {10}^{11}$

$E_1=\frac{12400}{4144}$ = 2.99 eV.

$E{­}_2=\frac{12400}{4972}$ = 2.49 eV.

$E_3=\frac{12400}{6216}$ = 1.99 eV.

Only the first two wavelengths are capable of ejecting photoelectrons.

Energy incident per second

= $\frac{3.6}{3}\times {10}^{-3}\times {10}^{-4}=1.2\times {10}^{-7}$ J /s

$n_1=\frac{1.2\times {10}^{-}7}{2.99\times 1.6\times {10}^{-19}}=2.5\times {10}^{11}$

$n_2=\frac{1.2\times {10}^{-7}}{2.99\times 1.6\times {10}^{-19}}=3\times {10}^{11}$

Total number of photons = 2 $(n_1+n_2)$

= 3.01$\times 1011+2.51\times 011=5.52\times {10}^{11}$

Total number of photo electrons ejected in two seconds = $11\times {10}^{11}$