Session #3: Homework Solutions

Transcription

1 Session #3: Homework s Problem #1 From a standard radio dial, determine the maximum and minimum wavelengths ( max and min ) for broadasts on the (a) AM band (b) FM band =, min = ; max = max min AM FM 3 x 10 m / s = = 1 m 1600 x 10 Hz min 3 3 x 10 = =.7 m 10 x 10 min 6 3 x 10 = = 566 m 530 x 10 max 3 3 x 10 = = 3.41 m x 10 max 6 Problem # For light with a wavelength ( ) of 40 nm determine: (a) the frequeny (b) the wave number () the wavelength in Å (d) the total energy (in Joules) assoiated with 1 mole of photons (e) the olor To solve this problem we must know the following relationships: = ; 1/ = ; 1 nm = 9 10 m = 10 Å E = h ; E = h x N (N = 6.0 x 10 ) molar A A 3 (a) (frequeny) = 3 x 10 m/s = = x 10 s x 10 m 14 1

2 1 1 (b) (wavenumber) = = =.45 x 10 m x 10 m 6 1 () Å = 40 x 10 m x = 400Å m (d) E = h x N = 6.63 x 10 x x 10 x 6.0 x 10 J/mole A = 5.93 x 10 J/mole = 93 kj/mole (e) visible spetrum: violet (500 nm) red (00 nm) 40 nm = UV Problem #3 For yellow radiation (frequeny,, = 5.09 x s 1 ) emitted by ativated sodium, determine: (a) the wavelength ( ) in [m] (b) the wave number ( ) in [m 1 ] () the total energy (in kj) assoiated with 1 mole of photons (a) The equation relating and is = where is the speed of light = 3.00 x 10 m x 10 m/s = = = 5.9 x 10 m x 10 s (b) The wave number is 1/wavelength, but sine the wavelength is in m, and the wave number should be in m 1, we first hange the wavelength into m: = 5.9 x 10 7 m x 100m/m = 5.9 x 10 5 m Now we take the reiproal of the wavelength to obtain the wave number: 1 1 = = = 1.70 x 10 m x 10 m 4-1 () The Einstein equation, E =h, will give the energy assoiated with one photon sine we know h, Plank s onstant, and. We need to multiply the energy obtained by Avogadro s number to get the energy per mole of photons. h = 6.6 x J.s 14-1 = 5.09 x 10 s E = h = (6.6 x 10 J.s) x (5.09 x 10 s ) = 3.37 x 10 J per photon

3 This is the energy in one photon. Multiplying by Avogadro s number: x 10 photons E N Av = (3.37 x 10 J per photon) mole =.03 x 10 5 J per mole of photons we get the energy per mole of photons. For the final step, the energy is onverted into kj: Problem #4 5 1 kj.03 x 10 J x =.03 x 10 kj 1000 J Potassium metal an be used as the ative surfae in a photodiode beause eletrons are relatively easily removed from a potassium surfae. The energy needed is.15 x 10 5 J per mole of eletrons removed (1 mole =6.0 x 10 3 eletrons). What is the longest wavelength light (in nm) with quanta of suffiient energy to ejet eletrons from a potassium photodiode surfae? anode + e I p h I p, the photourrent, is proportional to the intensity of inident radiation, i.e. the number of inident photons apable of generating a photoeletron. K layer This devie should be alled a phototube rather than a photodiode a solar ell is a photodiode. Required: 1eV = 1.6 x J E = h = (h)/ rad The question is: below what threshold energy (h ) will a photon no longer be able to generate a photoeletron? 5 1 mole.15 x 10 J/mole photoeletrons x 6.0 x 10 3 photoeletrons = 3.57 x J/photoeletron -34 h 6.6 x 10 x 3 x 10 = = = 5.6 x 10 m = 560 nm 3.57 x x 10 threshold

4 Problem #5 For red light of wavelength ( ) x 10 5 m, emitted by exited lithium atoms, alulate: (a) the frequeny ( ) in s 1; (b) the wave number ( ) in m 1 ; () the wavelength ( ) in nm; (d) the total energy (in Joules) assoiated with 1 mole photons of the indiated wavelength. (a) = and = / where is the frequeny of radiation (number of waves/s). For: -5 = x 10 m = x 10 m x 10 ms 14 1 = = x 10 s = Hz x 10 m (b) = = = x 10 m = x 10 m x 10 m 1 () -5 1 nm = x 10 m x = m 10 m (d) h 6.6 x 10 Js x.9979 x 10 ms E = = x 10 m = Problem # x 10 J/photon = 1.7 x 10 J/mole photons Calulate the Bohr radius for He +. In its most general form, the Bohr theory onsiders the attrative fore (Coulombi) between the nuleus and an eletron being given by: F = Ze 4πεor

5 where Z is the harge of the nuleus (1 for H, for He, et.). Correspondingly, the eletron energy (E el ) is given as: 4 Z me E el = - n h ε o and the eletroni orbit (r n ): n h εo n r = Z πme r = n n a Z o For He (Z=), r 1 = a o = x 10 m = 0.64Å Problem #7 (a) Determine the atomi weight of He ++ from the values of its onstituents. (b) Compare the value obtained in (a) with the value listed in your Periodi Table and 4 explain any disrepany if suh is observed. (There is only one natural He isotope.) (All relevant data are in the P/T and T/C.) (a) The mass of the onstituents (p + n) is given as: p = x x 10-4 g n = x x 10-4 g (p + n) = x 10-4 g The atomi weight (alulated) in amu is given as: x 10 g / amu x 10 g He = amu (b) The listed atomi weight for He is (amu). The data indiate a mass defet of

6 .941 x 10 amu, orresponding to 4.6 x 10 6 g/atom. This mass defet appears as nulear bond energy: ΔE = 4.6 x 10 kg x 9 x 10 m / s = x 10 J/atom =.6356 x 10 1 J/mole ΔE -5 Δm = =.9 x 10 kg/mole = 0.09 g/mole