Question

please explain c & d

11. The cutoff wavelength for the photoelectric effect in a certain metal is 254nm. a) What is the magnitude of the work function? b) Will the photoelectric effect be observed for wavelength greater than 254nm or for wavelength less than 254 nm? c) What is the kinetic energy of the electrons emitted if 300nm photons shine on this metal? d) What is the kinetic energy of the electrons emitted if photons of 200nm shine on this metal?

Answer 1

Solution: w we can write, where, he - Work function, wo a no his pland constant h = 6.62.681034 J. S . We - 6626 X16344.5 X3x108m 5- Do = Thresold frequency 254 * 10-9m do = wavelength (cut off) given, to = 254 nm = 7.825 X10-1951 = 254 x 10-9m : The magnitude of the C = velocity of light = 3x108mš Horw function is 7.825 xio 19 g Lins] b) Cut off wavelength is the maximum wavelength for which photoelectrie effect is observed. Above eat off wavelength the energy is too low to Knock of electrons. So, above cut off wavelength photo electric effect is not observed. Hence, the photo electric effect will not observed for wavelength greater than 254 nm. But, the photo electric effect is possible for ware length less than 254 nm. Icy for photo electric effect, We can write, the kinetic energy, E = NAMD = Nane MUY Where, NA is the Avogradro number, c is the velocity of light, & is the wavelength. - 30ohm = 300x109 m I hivena 1023 x 6.626 X 10 TS X 3 X 108 6.023 x 102 300 x10-9m = 395 X103 J ds Scanned with q uam canner

M And of the electrons emitted . So, the kinetic energy is 399x10²5 3oonm photons Shine on this metal. if Here, d = 200 nm = 200 x109m The Kinetic energy E = 6.023X102"x6.626x1094.5 x 3X10% m5! 200 X 10^9m = 598•62 X 103 y ans = The kinetic energy of the electrons emitted is 598.62 x 103 g if photons of 200 nm Shine on this metal. ds Scanned with CamScanner