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11.Dual Nature of Radiation and matter
hard
One milliwatt of light of wavelength $4560\ \mathring A $ is incident on a cesium surface of work function $1.9\ eV$. Given that quantum efficiency of photoelectric emission is $0.5\%$. Planck's constant $h = 6.62 \times 10^{-34}\ J-s$, velocity of light $= 3 \times 10^8\ ms^{-1}$, the corresponding photo electric current is
A$1.856 × 10^{-6}\ A$
B$1.856 × 10^{-7}\ A$
C$1.856 × 10^{-5}\ A$
D$1.856 × 10^{-4}\ A$
Solution
Photo current ${I_p} = {n_e}e$
where $\mathrm{n}_{\mathrm{e}}=$ no. of ${e^ – }s$ emitted
$ \mathrm{n}_{\mathrm{e}} =\frac{0.5}{100} \times \mathrm{n}_{\mathrm{p}} $
$=\frac{5}{1000} \times$ no. of photons falling the surface
$=\frac{5}{1000} \times\left(\frac{\mathrm{P} \lambda}{\mathrm{hc}}\right)$
$=\frac{5}{1000} \times 2.35 \times 10^{15}$
$=1.15 \times 10^{13}$
Therefore
${I_p} = {n_e}e = 1.15 \times {10^{13}} \times 1.6 \times {10^{ – 19}}$
$=1.856 \times 10^{-6} A$
where $\mathrm{n}_{\mathrm{e}}=$ no. of ${e^ – }s$ emitted
$ \mathrm{n}_{\mathrm{e}} =\frac{0.5}{100} \times \mathrm{n}_{\mathrm{p}} $
$=\frac{5}{1000} \times$ no. of photons falling the surface
$=\frac{5}{1000} \times\left(\frac{\mathrm{P} \lambda}{\mathrm{hc}}\right)$
$=\frac{5}{1000} \times 2.35 \times 10^{15}$
$=1.15 \times 10^{13}$
Therefore
${I_p} = {n_e}e = 1.15 \times {10^{13}} \times 1.6 \times {10^{ – 19}}$
$=1.856 \times 10^{-6} A$
Standard 12
Physics
