1. Photoelectric effect. Li metal is irradiated with light at two wavelengths. A

Question

1. Photoelectric effect. Li metal is irradiated with light at two wavelengths. At 300 nm the kinetic energy of the ejected electron is 2.935 x 1019 J. Alternatively, when irradiated at 400 nm the inetic energy for the ejected electron is 1.280 x 1019 J. Use this information to derive a value for (a) Plank's constant(h), the (b) threshold frequency (vo), and the work function (O) of lithium 2. Assume you want to use an electron microscope to make measurements on an atomic scale (a 250 nm). Relative to the speed of light (c), how fast would an electron need to be accelerated? (Calculate your answer in terms of(%) orfraction ofc)--ot-elecah, chon (rgk 3. 2D Particle in a box. As you know, a symmetric 2D particle in a box (Li-L2) results in degeneracy of energy levels. However, in some instances degenerate levels are encountered even if no rotation exists that can transform one wavefunction into another. This is referred to as ‘accidentally degeneracy. Conversely, situations were ‘hidden symmetry, exists, it is possible to transform one wavefunction into another by symmetry operations. a) Consider a rectangular box with sides Li -L and L2-2L. The potential energy within the box is zero and infinite outside the boundary of the 2D box. For this system, determine if there are any states which exhibit degeneracy for the first 6-quantum levels for x- and y- dimensions (n and ny). NOTE: A COMPUTER WOULD MAKE THIS EASY. b) Based on your knowledge of Group Theory, can any arguments be made regarding each instance of degeneracy observed? Specifically, try to provide an example for both 'accidental degeneracy' and 'degeneracy through hidden symmetry'. The electronic (UV-visible) spectrum for nitromethane (HiC-NO2) exhibits two maxima at 270 nm [molar extinction coefficient (e) ~ 20 M-1cmand 210 nm [e-16,000 M.,cm Using group theory to construct a qualitative molecular orbital scheme and assign the observed transitions. Make sure to specify for each transition if it is spin- and orbitally allowed (or forbidden) and explain your reasoning. HINT. FOR SIMPLICY, TREAT THE METHY GROUP AS A SPHERE AND ONLY CONSIDER THE SALC PRODUCED FROM THE NITRO GROUP. 4.

Explanation / Answer

We know that:

  hf = hc/ =   + Ek

hf (=hc/)= energy of irridiated electron(planck's constant x frequency, or planck's constant x speed of light/wavelenght of incident light)

= work function of metal

Ek = kinetic energy of ejected electron

1. at 300 nm = 300 x 10-9 m

h x 3 x 108 m/s / (300 x 10-9 m) = + 2.935 x 10-19J -----------(1)

2. at 400 nm

h x 3 x 108 m/s / 400 x 10-9 m = + 1.28 x 10-19 J -------------------(2)

subtracting (2) from (1)

h x 3 x 108 m/s / (300 x 10-9 m) - h x 3 x 108 m/s / 400 x 10-9 m = + 2.935 x 10-19J -( + 1.28 x 10-19 J )

h x 3 x 108 m/s / (100 x 10-9 m){1/3-1/4} = 1.655 x 10-19 J

h = 1.655 x 10-19 J x 12 x 100 x 10-9 m / 3 x 108 m/s = 6.62 x 10-34 J/s

putting this value in any of the two equation (1) or (2)

lets put the values in (1)

h x 3 x 108 m/s / (300 x 10-9 m) = + 2.935 x 10-19J

work function () = 3.685 x 10-19 J

hvo = 3.685 x 10-19 J

vo = threshold frequency

vo = 3.685 x 10-19 J / 6.62 x 10-34 J/s = 5.57 x 1014 s-1

2.

We know that :

= h/p de-broglie's equation

p = mv

m = mass of electron = 9.11 × 10-31 kg , v = velocity

250 nm = 250 x 10-9 m = 6.62 x 10-34 J/s / (9.11 × 10-31 kg x v )

v = 2906.69 m/s

% speed of light = speed / speed of light x 100 = 2906.69 m/s / 3 x 108 m/s = 9.69 x 10-4 speed of c %

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