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Light of wavelength $4000\: A^{\circ}$ and intensity $100 W/m^2$ is incident on a plate of threshold frequency $ 5.5 \times 10^{14}Hz$. The no. of photons incident per $m^2$ per second is

$\begin {array} {1 1} (a)\;2.02 \times 10^{20} & \quad (b)\;2.02 \times 10^{22} \\ (c)\;202 \times 10^{20} & \quad (d)\;202 \times 10^{22} \end {array}$

 

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Ans : (a)
$E_k = hv – hv_{\circ} = h[v – v_{\circ} ]$
$= h \bigg[\large\frac{c}{\lambda} – v_{\circ} \bigg]$
$= 6.6 \times 10^{-34} \times \bigg[ \large\frac{3 \times 10^8 }{4000 \times 10^{-10}} – 5.5 \times 10^{14 } \bigg] J$
$= 6.6 \times 10^{-34} [7.5 \times 10^{14} – 5.5 \times 10^{14} ] J$
$= 1.32 \times 10^{-19} J$
Let $n$ be no. of photons incident per $m^2$ per second, then $nhv = 100$
$ \Rightarrow \large\frac{nhc}{ \lambda} =100$
$ \Rightarrow n = \large\frac{100 \lambda}{hc}$
$ \Rightarrow n = \large\frac{100 \times 4000 \times 10^{-10}}{(6.6 \times 10^{-34} \times 3 \times 10^8 )} $$= 2.02 \times 10^{20}$
answered Jan 1, 2014 by thanvigandhi_1
edited Mar 25, 2014 by balaji.thirumalai
 

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