Modern Physics 7 Question 47
52. When a beam of $10.6 eV$ photons of intensity $2.0 W / m^{2}$ falls on a platinum surface of area $1.0 \times 10^{-4} m^{2}$ and work function $5.6 eV .0 .53 %$ of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per second and their minimum and maximum energies (in $eV$ ). Take $1 eV$ $=1.6 \times 10^{-19} J$.
$(2000,4 M)$
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Answer:
Correct Answer: 52. (i) A(ii) F (iii) E (iv) C
Solution:
- Energy of incident photons,
$$ \begin{aligned} E _i & =10.6 eV=10.6 \times 1.6 \times 10^{-19} J \\ & =16.96 \times 10^{-19} J \end{aligned} $$
Energy incident per unit area per unit time (intensity) $=2 J$
$\therefore$ Number of photons incident on unit area in unit time
$$ =\frac{2}{16.96 \times 10^{-19}}=1.18 \times 10^{18} $$
Therefore, number of photons incident per unit time on given area $\left(1.0 \times 10^{-4} m^{2}\right)$
$$ =\left(1.18 \times 10^{18}\right)\left(1.0 \times 10^{-4}\right)=1.18 \times 10^{14} $$
But only $0.53 %$ of incident photons emit photoelectrons
$\therefore$ Number of photoelectrons emitted per second (n)
$$ \begin{aligned} n & =\frac{0.53}{100}\left(1.18 \times 10^{14}\right) \\ n & =6.25 \times 10^{11} \\ K _{\min } & =0 \\ K _{\max } & =E _i-\text { work function } \\ & =(10.6-5.6) eV=5.0 eV \end{aligned} $$
$$ \begin{aligned} & \text { and } \quad K _{\max }=E _i \text {-work function } \\ & \therefore \quad K _{\max }=5.0 eV \\ & \text { and } \quad K _{\min }=0 \end{aligned} $$
