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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'.

Kapil Khare
80 Points
2 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’ = $\lambda$1 =200nm = 2*10^(-7)m
Energy of photon which hit plate ‘A’ = hc/$\lambda$1 = 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