Calculate the value of a) plank's constant and b) the work function of metal from the data obtained. Robert A millikan was attempting to disprove Einsteins photoelectric equation but found that his data supported Einsteins prediction.
Data(y = stopping potential (V); x = frequency (Hz x10^13)
a) y = 0 V; x = 43.9 hz x 10^13
b)y = 1 V; x = 70 hz x 10^13
c) y= 2V; x = 83hzx10^13
a) find h (planks constant) in J*s
b) find the work function(phi) in eV
Hint: threshold frequency f(o) is 43.9 hz x 10^13. Stopping potential is related to maximum kinetic energy of the ejected electrons by K(max) = e V(stopping potential)
Photoelectric equation from einstein is
k(max) = h*f- phi(work function)
V(s) = (h*f/e)-(phi/e)
The work function phi is given by the stopping potential V(o) = -phi/e
where V(s) = mf + V(o). V(s) = 0 when f = f(o)
Data(y = stopping potential (V); x = frequency (Hz x10^13)
a) y = 0 V; x = 43.9 hz x 10^13
b)y = 1 V; x = 70 hz x 10^13
c) y= 2V; x = 83hzx10^13
a) find h (planks constant) in J*s
b) find the work function(phi) in eV
Hint: threshold frequency f(o) is 43.9 hz x 10^13. Stopping potential is related to maximum kinetic energy of the ejected electrons by K(max) = e V(stopping potential)
Photoelectric equation from einstein is
k(max) = h*f- phi(work function)
V(s) = (h*f/e)-(phi/e)
The work function phi is given by the stopping potential V(o) = -phi/e
where V(s) = mf + V(o). V(s) = 0 when f = f(o)
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V(s) = (h*f/e) - (phi/e)
0 = (h*f/e) - phi/e when f = 43.9 x 10^13 hertz
so phi = h*(43.9 x 10^13)
1 V = (h*70x 10^13/e) - (h* 43.9 x 10^13/e)
1 eV = h(26.1 x 10^13 hertz)
e = 1.602 x 10^-19 Coulombs
h = (1.602 x 10^ -19 Coulomb Volt second)/26.1 x 10^13
h = 6.138 x 10^-34 joule seconds since a coulomb volt is a joule
since phi = (43.9 x 10^13)h and 1 eV = (26.1 x 10^13)h
phi = (43.9 x 10^13) (1eV/26.1 x 10^13) = 1.682 eV
0 = (h*f/e) - phi/e when f = 43.9 x 10^13 hertz
so phi = h*(43.9 x 10^13)
1 V = (h*70x 10^13/e) - (h* 43.9 x 10^13/e)
1 eV = h(26.1 x 10^13 hertz)
e = 1.602 x 10^-19 Coulombs
h = (1.602 x 10^ -19 Coulomb Volt second)/26.1 x 10^13
h = 6.138 x 10^-34 joule seconds since a coulomb volt is a joule
since phi = (43.9 x 10^13)h and 1 eV = (26.1 x 10^13)h
phi = (43.9 x 10^13) (1eV/26.1 x 10^13) = 1.682 eV
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