The threshold wavelength of tungsten is 2.76 x 10-5 cm. (a) Explain why no photoelectrons are emitted when the wavelength is more than 2.76 x 10-5 cm. (b) What will be the maximum kinetic energy of electrons ejected in each of the following cases (i) if ultraviolet radiation of wavelength a = 1.80 x 10-5 cm and (ii) radiation of frequency 4x1015 Hz is made incident on the tungsten surface.
Answers
λ0 = 2.76 × 10-5 cm = 2.76 × 10-7 m, λ =1.80 × 10-5 cm = 1.80 × 10-7 m, v = 4 × 1015 Hz, h = 6.63 × 10-34 J∙s,c = 3 × 108 m/s (a) For λ > λ0 , v < v0 (threshold frequency). ∴ hv < hv0 . Hence, no photoelectrons are emitted. (b) Maximum kinetic energy of electrons ejected = hc ( 1 λ − 1 λ o ) (1λ−1λo) = (6.63 × 10-34 )(3 × 108) ( 10 7 1.8 − 10 7 2.76 ) (1071.8−1072.76)J = (6.63 × 10-19) (0.5555 – 0.3623) = (6.63)(0.1932 × 10-19) J = 1.281 × 10-19 J = 1.281 × 10 − 19 J 1.6 × 10 − 19 J e V 1.281×10−19J1.6×10−19JeV = 0.8006 eV (c) Maximum kinetic energy of electrons ejected = hv – h c λ o hcλo =(6.63 × 10-34 (4 × 1015 ) – ( 6.63 × 10 − 34 ) ( 3 × 10 8 ) 2.76 × 10 − 7 (6.63×10−34)(3×108)2.76×10−7 = 26.52 × 10-19 – 7.207 × 10-19 = 19.313 × 10-19 J = 19.313 × 10 − 19 J 1.6 × 10 − 19 J e V 19.313×10−19J1.6×10−19JeV = 12.07eVRead more on Sarthaks.com - https://www.sarthaks.com/1371093/the-threshold-wavelength-of-tungsten-is-2-76-10-5-cm