11.11 Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.

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Payal Gupta | Contributor-Level 10

8 months ago

11.11 Wavelength of the light produced by Argon laser, $λ$ = 488 nm = 488 $* 10^{- 9}$ m

Stopping potential of the photoelectrons, $V_{0}$ = 0.38 V = $\frac{0.38}{1.6 * 10^{- 19}}$ $e V$

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

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

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

From Einstein's photoelectric effect, we have the relation involving the work

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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)
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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

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$v_{1}^{2} = \frac{2 h υ_{0}}{m}$

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

Sol:

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