Orange light with a wavelength of 6.00x102nm is directed at a metallic surface with a work function of 1.60eV. Calculate:

a) The maximum kinetic energy of the emitted electrons
b) Their maximum speed
c) The cutoff potential necessary to stop these electrons (V=Ek/e)

wait...did u mean a wavelength of "6.00 x 10^2"?

if so...these are my answers

a) first convert the numbers

6.00 x 10^2 nm = 6 x 10^-7 m
1.60 eV = 2.56 x 10^-19

then, plug in the numbers in this equation:

KE = hc/l - w

KE = (6.63E-34)(3E8)/(6E-7) - (2.56E-19) = 7.55E-20 J

b) for the velocity, use the ususal kinetic energy equation and solve for velocity

KE = (1/2)mv^2

then...

v = squ-root(2 x KE/m)

(assuming the mass is an electron...)

v = squ-root(2 x (7.55E-20)/(9.1094E-31)) = 4.0714E5 m/s

c) i have no idea how to do...srry bout that mate...

To find the answers to these questions, we can use the concept of the photoelectric effect and the equations associated with it.

The photoelectric effect is the phenomenon in which electrons are emitted from a metal surface when light of a certain frequency, or equivalently, wavelength, is incident upon it. The energy of the incident photons must be equal to or greater than the work function of the metal in order for electrons to be emitted.

a) The maximum kinetic energy (Ek) of the emitted electrons can be calculated using the equation: Ek = hf - φ, where h is the Planck's constant (6.63 x 10^-34 J·s) and f is the frequency of the incident light.

To calculate f from the given wavelength (λ), we can use the relationship between frequency and wavelength, which is given by: c = λf, where c is the speed of light (3.00 x 10^8 m/s). Rearranging the equation, f = c/λ.

Substituting the values into the equation Ek = hf - φ:
Ek = (hc/λ) - φ
= (6.63 x 10^-34 J·s * 3.00 x 10^8 m/s) / (6.00 x 10^2 nm) - 1.60 eV

First, we need to convert the wavelength from nm to meters. Since 1 nm = 1 x 10^-9 m, the wavelength becomes:
λ = 6.00 x 10^2 nm = 6.00 x 10^-9 m

Next, we need to convert the work function from eV to joules. Since 1 eV = 1.6 x 10^-19 J, the work function becomes:
φ = 1.60 eV = 1.60 x 10^-19 J

Substituting the values:
Ek = (6.63 x 10^-34 J·s * 3.00 x 10^8 m/s) / (6.00 x 10^-9 m) - 1.60 x 10^-19 J

Now, calculate Ek.

b) The maximum speed of the emitted electrons can be calculated using the equation: v = √(2Ek/m), where v is the speed, Ek is the kinetic energy (calculated in part a), and m is the mass of an electron (9.11 x 10^-31 kg).

Substituting the values:
v = √((2 * Ek) / m)

c) The cutoff potential (V) necessary to stop these electrons can be calculated using the equation: V = Ek/e, where e is the elementary charge (1.60 x 10^-19 C).

Substituting the values:
V = Ek / e

Once you calculate the values for these equations, you will have the answers to all three questions.