11.7 A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm.

(a) What is the energy per photon associated with the sodium light?

(b) At what rate are the photons delivered to the sphere?

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Payal Gupta | Contributor-Level 10

8 months ago

11.7 Power of the sodium lamp, P = 100 W

Wavelength of the emitted sodium light, $λ$ = 589 nm = 589 $* 10^{- 9}$ m

Planck's constant, h = 6.626 $* 10^{- 34}$ Js

Speed of light, c = 3 $* 10^{8}$ m/s

Energy per photon associated with the sodium light is given as:

$E$ = $\frac{h c}{λ}$ = $\frac{6.626 * 10^{- 34} * 3 * 10^{8}}{589 * 10^{- 9}}$ = 3.37 $* 10^{- 19}$ J = $\frac{3.37 * 10^{- 19}}{1.6 * 10^{- 19}}$ eV = 2.11 eV

Let the number of photon delivered to the sphere

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$_{} \frac{}{}$ $K . E_{1} = \frac{1230}{800} - ?$ - (i)

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hu = hu₀ + K.E

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h2u₀ = hu₀ + K.E₁

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$v_{1}^{2} = \frac{2 h υ_{0}}{m}$

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$\frac{v_{2}}{v_{1}} = \sqrt[]{\frac{8}{2} = 2}$

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Sol:

$\begin{array}{l} f (x) = \frac{2 x^{2} - 3 x - 2}{x - 2} \\ = \frac{2 x^{2} - 4 x + x - 2}{x - 2} = \frac{2 x (x - 2) + 1 (x - 2)}{x - 2} \\ = \frac{(2 x + 1) (x - 2)}{x - 2} = 2 x + 1 \\ \underset{x \to 2^{-}}{l i m} f (x) = 2 x + 1 = \underset{h \to 0}{l i m} 2 (2 - h) + 1 = 4 + 1 = 5 \\ \underset{x \to 2^{+}}{l i m} f (x) = 2 x + 1 = \underset{h \to 0}{l i m} 2 (2 + h) + 1 = 4 + 1 = 5 \\ \underset{x \to 2}{l i m} f (x) = 5 \\ A s \underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2}{l i m} f (x) = 5 \\ H e n c e, f (x) i s c o n t i n u o u s a t x = 2 . \end{array}$