Electrons are ejected from a metallic surface with speeds of up to 4.60 3 105 m/s when light with a wavelength of 625 nm is used. (a) What is the work function of the surface? (b) What is the cutoff f

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?

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

the answer is a. at the equator

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6