I WB	Temat: Wyznaczenie stałej Plancka oraz pracy wyjścia elektronu	23.04.2008
Nr cw. 9	Marek Nalepka

Nr filtru	Długośc fali λ [nm]	Szerokość połówkowa τ [nm]	Częstotliwość fotonu	napięcie hamujące [mV]
	[nm]	[nm]	[1/s]	Uśr
1	368	10	8,152	949
2	373	10	8,043	930
3	405	20	7,407	767
4	417	6	7,194	641
5	428	10	7,009	581
6	440	10	6,818	536
7	445	10	6,745	490
8	452	10	6,637	451

Częstotliwość fotonu $\nu = \frac{c}{\lambda}\ \left\lbrack \frac{1}{s} \right\rbrack$

c – prędkość światła

λ − dlugosc fali 

$v_{1} = \frac{300000000}{0,000000368} = 8,152$ v₂ = 8, 043 v₃ = 7, 407 v₄ = 7, 194 v₅ = 7, 009

v₆ = 6, 818 v₇ = 6, 742 v₈ = 6, 637

Obliczam $\frac{1}{\lambda}$

$\frac{1}{\lambda_{1}} = 2,717 \times 10^{6}\ $ $\frac{1}{\lambda_{2}} = 2,681 \times 10^{6}$ $\frac{1}{\lambda_{3}} = 2,469 \times 10^{6}$ $\frac{1}{\lambda_{4}} = 2,398 \times 10^{6}$

$\frac{1}{\lambda_{5}} = 2,336 \times 10^{6}$ $\frac{1}{\lambda_{6}} = 2,272 \times 10^{6}$ $\frac{1}{\lambda_{7}} = 2,247 \times 10^{6}$ $\frac{1}{\lambda_{8}} = 2,212 \times 10^{6}$

Współczynnik a i b za pomocą regresji liniowej

$$a = \left\lbrack 8 \times \sum_{i = 1}^{8}{U_{i}\left( \frac{1}{\lambda} \right)_{i}} - \left( \sum_{i = 1}^{8}U_{i} \right)\left( \sum_{i = 1}^{8}\left( \frac{1}{\lambda} \right)_{i} \right) \right\rbrack \times \frac{1}{X}$$

$$b = \left\lbrack \left( \sum_{i = 1}^{8}\left( \frac{1}{\lambda} \right)_{i}^{2} \right)\left( \sum_{i = 1}^{8}U_{i} \right) - \left( \sum_{i = 1}^{8}\left( \frac{1}{\lambda} \right)_{i} \right)\left( \sum_{i = 1}^{8}{U_{i}\left( \frac{1}{\lambda} \right)_{i}} \right) \right\rbrack \times \frac{1}{X}$$

$$X = 8\left( \sum_{i = 1}^{8}\left( \frac{1}{\lambda} \right)_{i}^{2} \right) - \left( \sum_{i = 1}^{8}\left( \frac{1}{\lambda} \right)_{i} \right)^{2}$$

$$\sigma = \sqrt{\frac{\sum_{i = 1}^{n}\left( U_{i} - a\left( \frac{1}{\lambda} \right)_{i} - b \right)^{2}}{n - 2}}$$

$$S_{a} = \sigma\sqrt{\frac{n}{X}}$$

$$S_{b} = \sigma\sqrt{\frac{\sum_{i = 1}^{8}\left( \frac{1}{\lambda} \right)_{i}^{2}}{X}}$$

$$\sum_{i = 1}^{8}U_{i} = 5,345$$

$$\sum_{i = 1}^{8}\left( \frac{1}{\lambda} \right)_{i} = 19,334\ \times 10^{6}\ $$

$$\sum_{i = 1}^{8}U_{i}^{2} = 3,833$$

$$\sum_{i = 1}^{8}\left( \frac{1}{\lambda} \right)_{i}^{2} = 46,988 \times 10^{12}$$

$$\sum_{i = 1}^{8}{U_{i}\left( \frac{1}{\lambda} \right)_{i}} = 13,178\ \times 10^{6}$$

X = 1, 267  × 10¹²

a = 0, 849 x 10 ^-6

b  =  1, 537 

σ = 3240

S_a = 0, 012 × 10⁻⁶

S_b= 0,38

Dla sprawdzenia obliczam współczynnik regresji liniowej $tga = \frac{U}{\frac{1}{\lambda}}$

$$tga = \frac{U}{\frac{1}{\lambda}} = \frac{431}{468,5} = 0,9198 \times 10^{- 6}$$

Stałą Plancka $h = \frac{a \times e}{c}$

$$h\ = \ \frac{0,849 \times 10^{- 6} \times 1,6 \times 10^{- 19}}{3 \times 10^{8}} = {5,932}^{- 34}$$

$$u\left( h \right) = \sqrt{\left( \frac{1,6\ x\ 10^{- 19}}{{3\ x\ 10}^{8}}\ x\ 0,012\ x\ 10^{- 6} \right)^{2}} = {6,23}^{- 36}$$

Niepewność u(U)= U×0,05

U₁ = 0, 949 × 0, 05 = 0, 047 U₂ = 0, 046 U₃ = 0, 039 U₄ = 0, 032 U₅ = 0, 029 U₆ = 0, 026 U₇ = 0, 024 U₈ = 0, 023

Niepewność $u(\lambda) = \ \frac{\tau}{2}$

$u\left( e\lambda_{2} \right) = \ \frac{10}{2} = 5nm$ u(λ₃)=10nm u(λ₄)=3nm u(λ₅)=5nm

u(λ₆)=5nm u(λ₇)=5nm u(λ₈)=5nm

Niepewność dla $\frac{1}{\lambda} = \ \sqrt{\left\lbrack \frac{1}{\lambda_{i} \times 10^{- 9}} \times u(\lambda) \times 10^{- 9} \right\rbrack^{2}}$

$\frac{1}{u\left( \lambda_{2} \right)} = 0,01$ $\frac{1}{u\left( \lambda_{3} \right)} = 0,03$ $\frac{1}{u\left( \lambda_{4} \right)} = 0,01$ $\frac{1}{u\left( \lambda_{5} \right)} = 0,01$ $\frac{1}{u\left( \lambda_{6} \right)} = 0,01$

$\frac{1}{u\left( \lambda_{7} \right)} = 0,01$ $\frac{1}{u\left( \lambda_{8} \right)} = 0,01$

Praca wyjścia elektronu ze wzoru W = e × b

b=1,537

$\overset{\overline{}}{W} =$ 1, 537 × 1, 6  × 10⁻¹⁹ = 2, 46 × 10⁻¹⁹J × s

$$u\left( W \right) = \sqrt{\left( \text{e\ x\ }S_{b} \right)^{2}}$$

$$u\left( W \right) = \sqrt{\left( 1,6\ x\ 10^{- 19}\ x\ 0,38 \right)^{2} =}4,76\ x\ 10^{- 20}$$