15.The de Broglie wavelength of electron of He ion is 3.329 Å. If photon emitted upon de-excitationof this He* ion is made to hit H atom in its ground state so as to liberate electron from it, what willbe the de Broglie's wavelength of photoelectron?
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C)
total energy = -13.58eV
En=-13.6 x (Z2/n2)eV
For He+ion, Z=2 and En = -13.58eV
So, n2= -13.6x4 / -13.58 = n = 2.
Thus, He+ion is in the 1 excited state.
Energy of photon emitted
Energy of photon =I.P. of H + K.E.of photoelectron Thus, K.E. of photoelectron= 40.8-13.6 = 27.2eV
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Answer:
2.3518
Explanation:
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