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
Answer:
(a) 75ev
(b) 1.414×10^(-10)m
(c) 2×10^-14 m
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,
(1)
Where,
Ф=striking energy
Ф₀=wave function
=maximum kinetic energy
The striking energy(Ф),
(2)
Inserting values in equation (2) we get;
(3)
By inserting the values in equation (1) we get;
(4)
(b)
The wavelength of the electron with maximum kinetic energy (in Å) is,
(5)
The wavelength of electrons with maximum kinetic energy (in Å) is 165.3.
(c) The uncertainty in wavelength is calculated as,
(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;
The uncertainty in the momentum is 7.9×10⁻⁸m.
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