Light of wavelength 2000 A falls on an aluminium surface (work function of aluminium 4.2 eV). Calculate
(a) the kinetic energy of the fastest and slowest emitted photoelectrons
(b) stopping potential
(c) cut-off wavelength for aluminium.

Solution

(a)
We know that,
Energy of photon =

\frac{hc}{λ}

\Rightarrow

E = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{2000 \times 10^{- 10}} J

i.e.,

E = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{2 \times 10^{- 7} \times 1.6 \times 10^{- 19}} eV = 6.20 eV

Energy of the fastest emitted photoelectron,
= h (v – v₀) (Where v₀ is the work function)
= (6.2 – 4.2) eV
= 2.0 eV

Since, the emitted electrons from a metal surface have an energy distribution, the minimum energy in this distribution being zero, the energy of slowest photoelectrons is also zero.

(b) Since, ${eV}_{0} = \frac{1}{2} {mv}^{2}_{\max}$
where, is the maximum energy of the emitted photo electrons and V₀ is the stopping potential, the stopping potential is 2V.

(c) The threshold frequency is related to the work function by the relation,
$ϕ_{0} = {hv}_{0} = \frac{hc}{λ_{0}}$

$∴$ $\frac{1}{λ_{0}} = \frac{ϕ_{0}}{hc} = \frac{4.2 \times 1.6 \times 10^{- 19} J}{6.62 \times 10^{- 34} Js \times 3 \times 10^{8} {ms}^{- 1}}$

$∴$ $λ_{0} = 3 \times 10^{- 7} m = 3000 Å$ , is the required cut-off wavelength for aluminum.

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Wave Optics

Light of wavelength 2000 A falls on an aluminium surface (work function of aluminium 4.2 eV). Calculate
(a) the kinetic energy of the fastest and slowest emitted photoelectrons
(b) stopping potential
(c) cut-off wavelength for aluminium.

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Wave Optics

Light of wavelength 2000 A falls on an aluminium surface (work function of aluminium 4.2 eV). Calculate(a) the kinetic energy of the fastest and slowest emitted photoelectrons(b) stopping potential(c) cut-off wavelength for aluminium.

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Light of wavelength 2000 A falls on an aluminium surface (work function of aluminium 4.2 eV). Calculate
(a) the kinetic energy of the fastest and slowest emitted photoelectrons
(b) stopping potential
(c) cut-off wavelength for aluminium.