11.9 The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?

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Payal Gupta | Contributor-Level 10

8 months ago

11.9 Work function of the metal, $\emptyset_{0}$ = $4.2 e V$

Charge of an electron, e = 1.6 $* 10^{- 19}$ C

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

Wavelength of the incident radiation, $λ$ = 330 nm= 330 $* 10^{- 9}$ m

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

The energy of the incident photon is given as:

$E$ = $\frac{h c}{λ}$ = $\frac{6.626 * 10^{- 34} * 3 * 10^{8}}{330 * 10^{- 9}}$ = 6.02 $* 10^{- 19}$ J = $\frac{6.02 * 10^{- 19}}{1.6 * 10^{- 19}}$ eV = 3.76 eV

Since the energy of the incident pho

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Radiation from hydrogen gas excited to first excited state is used for illuminating certain metallic plate. When the same plate is exposed to the radiation from some unknown hydrogen like gas excited to the same level it is found that de-Broglie wavelength of fastest photoelectron has decreased by 2.3 times. It is given that energy corresponding to longest wavelength of Lyman series of the unknown gas is 3 times the ionization energy of hydrogen gas (13.6eV). The work function (in eV) of the metallic plate is ______. [Take (2.3)² = 5.25]

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z² × (13.6) (1 - ¼) = 3 × (13.6)
z = 2 . (i)
h/√2mk? = (1/2.3) × h/√2mk?
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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

K.E₁ = hu₀

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$v_{1}^{2} = \frac{2 h υ_{0}}{m}$

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h 5u₀ = hu₀ + k.E₂

k.E₂ = 4hu₀

$\frac{1}{2} m v_{2}^{2} = 4 h υ_{0}$

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

v₂ = 2v₁

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This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

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