Complete question:
Electrons are ejected from a metallic surface with speeds ranging up to 4.60 x10⁵ m/s when light with a wavelength of 625nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface?
Answer:
Part(a) The work function of the surface is 22.177 x 10⁻²⁰ J = 1.384 eV
Part(b) The cutoff frequency for this surface is 3.347 x 10¹⁴ Hz
Explanation:
The kinetic energy (KE) of the emitted photon:
KE = 0.5mv²
m is mass of electron = 9.1 X 10⁻³¹ kg
KE = 0.5 * 9.1 X 10⁻³¹ * (460000)² = 9.628 X 10⁻²⁰ J
in eV = 9.628 X 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 0.601 eV
The photon energy of the incoming radiation:
E = hf = hc/λ
c is speed of light (photon) = 3 x 10⁸
h is Planck's constant = 6.626 × 10⁻³⁴ J.s
E = (6.626 × 10⁻³⁴ *3 x 10⁸) /(625 X 10⁻⁹)
E = 31.805 X 10⁻²⁰ J
in eV = 31.805 X 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 1.985 eV
Part (a) the work function of the surface
KE = hf - W
where;
W is work function
W = hf - KE
W = 31.805 X 10⁻²⁰ J - 9.628 X 10⁻²⁰ J = 22.177 x 10⁻²⁰ J
in eV = 22.177 x 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 1.384 eV
Part(b) the cutoff frequency for this surface
W =hf
f = W/h
f = (22.177 x 10⁻²⁰ J)/(6.626 × 10⁻³⁴ J.s)
f = 3.347 x 10¹⁴ Hz
