Light of wavelength $2200 A^{o}$ (angstorm) falls on a photosensitive plate with work function $4.1$ eV. Find (a) energy of photon in eV (electron volt). (b) maximum kinetic energy of photoelectron and (c) stopping potential.

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Solution

Energy of photon= $\frac{h c}{λ} j o u l e s$
$λ = 2200 A^{0}$
Work function $ϕ = 4.1 e V$
$1 e V = 1.6 \times 10^{- 19} j o u l e s$

(a) $E = \frac{12375}{λ (A^{\circ})} e V$

$E = 5.6 e V$

(b)maximum Kinetic energy $K E_{m a x} = E - ϕ$
$K E_{m a x} = 5.6 - 4.1 = 1.5 e V$

(c)stopping potential= $\frac{K E_{m a x}}{e}$

stopping potential $= 1.5 V$

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Light of wavelength 2200Ao (angstorm) falls on a photosensitive plate with work function 4.1 eV. Find (a) energy of photon in eV (electron volt). (b) maximum kinetic energy of photoelectron and (c) stopping potential.

Energy of photon=hcλjoulesλ=2200A0Work function ϕ=4.1eV1eV=1.6×10−19joules(a) E=12375λ(A∘)eVE=5.6eV(b)maximum Kinetic energy KEmax=E−ϕKEmax=5.6−4.1=1.5eV(c)stopping potential=KEmaxestopping potential=1.5V

Light of wavelength $2200 A^{o}$ (angstorm) falls on a photosensitive plate with work function $4.1$ eV. Find (a) energy of photon in eV (electron volt). (b) maximum kinetic energy of photoelectron and (c) stopping potential.