Chemistry, asked by Narendrad42151, 6 months ago

Photon having wavelength 12.4 nm was allowed to strike a metal plate having work function 25 eV. Calculate the (a) Maximum kinetic energy of photoelectrons emitted (in eV). (b) Wavelength of electron with maximum kinetic energy (in Å). (c) Calculate the uncertainty in wavelength of emitted electron if the uncertainty in the momentum is 6.62 × 10–28 Kg m/sec​

Answers

Answered by visshu8853
1

Answer:

(a) 75ev

(b) 1.414×10^(-10)m

(c) 2×10^-14 m

Answered by archanajhaasl
0

Answer:

(a)The maximum kinetic energy of photoelectrons emitted (in eV) is 75.

(b)The wavelength of electrons with maximum kinetic energy (in Å) is 165.3.

(c)The uncertainty in the momentum is 7.9×10⁻⁸m.

Explanation:

From the question we have,

The wavelength which is striking the metal(λ)=12.4nm=124×10⁻¹⁰m

The work function of the metal(Ф₀)=25eV

(a)

From the photoelectric equation we have,

\mathrm{\Phi=\Phi_o+KE_{max}}      (1)

Where,

Ф=striking energy

Ф₀=wave function

\mathrm{KE_{max}}=maximum kinetic energy

The striking energy(Ф),

\mathrm{\Phi=\frac{12400 A^{\circ}}{\lambda} }           (2)

Inserting values in equation (2) we get;

\mathrm{\Phi=\frac{12400 A^{\circ}}{124 A^{\circ}} }

\mathrm{\Phi=100\ eV}         (3)

By inserting the values in equation (1) we get;

\mathrm{100=25+KE_{max}}

\mathrm{KE_{max}=75 \ eV}         (4)

(b)

The wavelength of the electron with maximum kinetic energy (in Å) is,

\mathrm{\lambda=\frac{12400\ A^{\circ}}{75\ eV}=165.3\ A^{\circ} }       (5)

The wavelength of electrons with maximum kinetic energy (in Å) is 165.3.

(c) The uncertainty in wavelength is calculated as,

\mathrm{\lambda=\frac{h}{4\pi \Delta P} }               (6)

Where,

λ=uncertanity in wavelength

h=planck's consatnt=6.62×10⁻³⁴Joule-second

ΔP=uncertanity in momentum=6.626×10⁻²⁸Kg m/sec​

Inserting the required values in equation (6) we get;

\mathrm{\lambda=\frac{6.62\times 10^{-34}}{4\pi\times  6.62\times 10^{-28}} }

\mathrm{\lambda=\frac{10^{-6}}{4\pi} }

\mathrm{\lambda=0.079\times 10^{-6}=7.9\times 10^{-8}\ m}

The uncertainty in the momentum is 7.9×10⁻⁸m.

#SPJ2

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