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# When a certain metal was irradiated with light of frequency 3.2×1016 Hz, the photoelectrons emitted had twice the kinetic energy as that of the photoelectrons emitted when the same metal was irradiated with light of frequency 2.0×1016 Hz. The value of ν∘ for the metal was found to be x×1016 Hz. Then x=

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Solution

## According to photoelectric effect: hν = w + K.E. where, h = Planck's constant hν = Energy of photon w = Work function = hνo νo = threshold frequency K.E. = Kinetic energy of electrons released from the metal surface As per the given statements, h ( 3.2 ×1016)=w+2K.E ........(1) h ( 2×1016)=w+K.E. ............(2) Subtracting (2) from (1) we get, h (1.2 ×1016) =K.E. 6.626×10−34×1.2×1016=7.95×10−18=K.E. Put value of K.E in equation (2) h ( 2×1016)=hνo+K.E. (6.626×10−34×2×1016)=hν∘ +7.95×10−18 13.25×10−18−7.95×10−186.626×10−34 = ν∘ On solving, νo = 7.99×1015 = 0.8×1016 Hz

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