Question
Download Solution PDFThe photoelectric threshold wavelength of silver is 3250 × 10–10 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 × 10–10 m is
(Given h = 4.14 × 10–15 eVs and c = 3 × 108 ms–1)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Photoelectric effect is the phenomenon of ejection of electrons from metal surface when light of frequency greater than the threshold frequency of the metal surface electrons.
This process is also known as photoemission and the electrons that are ejected from the metal are called photoelectrons.
The energy of photon is related to its frequency by relation:
\(E=h\nu=\frac{hc}{\lambda}\)
where h is Planck's constant, \(\nu\) is the frequency of light, c is velocity of photon, \(\lambda\) is wavelength of the photon.
The relationship between the Kinetic energy and energy of the incident photon is :
Ep = Threshold energy \((\phi)\) + Ee
where Ep and Ee are energy of incident photon and energy of emitted electron respectively.
\(h\nu= h\nu_o +\frac{1}{2}mv^2\)
where \(\nu_o\) is the threshold frequency of ejected electron, m is mass and v is velocity of electron.
Calculation:
λ0 = 3250 × 10-10 m
λ = 2436 × 10-10 m
\(\phi = \frac{1242 eV -nm}{324 \ nm} = 3.82 eV\)
\(\rm hv = \frac{1242 \ eV -nm}{253.6 \ nm}=4.89 \ eV\)
KEmax = (4.89 - 3.82) eV = 1.077 eV
\(\rm\frac{1}{2}mv^2\) \(=1.077 \times 1.6\times 10^{-19}\)
\(\rm v = \sqrt {\frac{{2\times 1.077 \times 1.6\times10^{-19}}}{9.1 \times 10^{-31}}} \)
\(\rm v = 0.6 \times 10^6 m/s\)
Last updated on Jun 16, 2025
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