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The electromagnetic radiation of wavelength 200 nm are incident on a metal plate 'A'. The photo electrons are accelerated by potential difference 10.2eV. These electron strike another metal plate 'B' from which electromagnetic radiation are emitted. The minimum wavelength of the emitted Photons is 100nm. Find the work function of the metal 'A'.

User , 6 Years ago
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Kapil Khare

Last Activity: 6 Years ago

Given that, minimum wavelength of the emitted photon = \lambda= 100nm = 10^(-7)m
 
Maximum energy received by electron = hc/\lambda = 12.40eV
\implies  Maximum energy of the photoelectron when it hit metal plate ‘B’ = 12.40eV
 
We know that,
Energy of electron when it left plate ‘A’ + Potential energy through which it is accelerated = Energy of electron when it hit plate ‘B’
\implies  K.Emax + 10.2 = 12.4
\implies  K.Emax = 2.2eV
 
Wavelength of photon which hit plate ‘A’ = \lambda1 =200nm = 2*10^(-7)m
Energy of photon which hit plate ‘A’ = hc/\lambda1 = 6.20eV
 
Energy of photon – Work function of plate A = Maximum Kinetic energy of electron
\implies 6.2 – W = 2.2
\implies W = (6.2 – 2.2)eV
\implies W = 4.0eV

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