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Light of frequency 7.21 x 10^4 Hz is incident on a metal surface. The cut-off wavelength for photoelectric emission from the metal surface is 540 nm. Determine the maxim n speed of thephotoelectrons emitted from the surface. (Given: h= 6.63 x 10-34 Js, mass of an electron = 9.1 x 10-31kg). |
| Answer» EXPLANATION:Given frequency of LIGHT= 7.21*10^4 Hzcut off WAVELENGTH= 540 nmKE(max)= 1/2mv^2= h(v-v0)v0= v- MV^2/2hV0= 7.21- 9.1*10^-31 * (3*10^8)^2/2(6.63*10^-34)V0= 4.74* 10^14 HzNote: in the given question, speed should be given , and we NEED to find threshold frequency. | |