Monochromatic light of an unknown wavelength passes through a single slit of width 5.59 μm and forms an interference pattern on a screen 1.26 m away.  The width of the central bright fringe of the interference pattern is 18.8 cm.  The same light is then shone on potassium, a metal with a work function of 2.28 eV.

 
a)  What is the wavelength of the light in nanometers?

b)  What is the maximum speed of the electrons emitted by the metal?

Warning:  Retain many sig digs in intermediate results.  Use the accepted values of all physical constants and conversion factors accurate to three sig digs, e.g. h = 6.63e-34 Js.

To find the wavelength of the monochromatic light, we can use the formula for the interference pattern:

λ = (m * λ * D) / w

where λ is the wavelength of the light, m is the order of the fringe, D is the distance between the screen and the slit, and w is the width of the single slit.

Given values:
m = 0 (for the central bright fringe)
D = 1.26 m
w = 5.59 μm (convert to meters: 5.59 * 10^-6 m)

a) Using the given values, we can rearrange the formula to solve for λ:

λ = (m * λ * D) / w
0 = (λ * 1.26) / 5.59 * 10^-6

Simplifying and solving for λ:
λ = (0 * 5.59 * 10^-6) / 1.26
λ = 0 nm

The wavelength of the light is 0 nm. (Please note that this answer may be incorrect or indicate a special case, as the given values lead to an invalid result.)

b) To find the maximum speed of the electrons emitted by potassium, we can use the formula for the maximum kinetic energy of the photoelectrons:

KEmax = h * ν - φ

where KEmax is the maximum kinetic energy, h is the Planck's constant (6.63 * 10^-34 Js), ν is the frequency of the light, and φ is the work function of potassium (2.28 eV).

To find the frequency of the light, we can use the equation:

c = ν * λ

where c is the speed of light (3 * 10^8 m/s).

By rearranging the equation, we can solve for ν:

ν = c / λ
ν = (3 * 10^8 m/s) / (0 nm) (using the incorrect wavelength value previously calculated)

Unfortunately, we cannot find the frequency of the light without a valid wavelength value. As a result, we cannot calculate the maximum speed of the electrons emitted by the metal.

To find the wavelength of the light, we can use the single-slit diffraction formula:

λ = (d * sinθ) / m

where λ is the wavelength, d is the width of the slit, θ is the angle between the central maximum and the mth order bright fringe, and m is the order of the bright fringe.

In this case, we are given the width of the slit (d = 5.59 μm), the distance to the screen (L = 1.26 m), and the width of the central bright fringe (W = 18.8 cm). We can convert W to meters by dividing by 100:

W = 18.8 cm = 0.188 m

The angle θ can be approximated as the tangent of the angle:

θ ≈ (0.188 m) / (1.26 m) = 0.1492 radians

Since this is a single-slit experiment, we are looking at the first-order bright fringe, so m = 1.

Now we can rearrange the formula to solve for the wavelength:

λ = (d * sinθ) / m
= (5.59 μm * sin(0.1492)) / 1

Calculating this, we find:

λ ≈ 0.01351 μm

To convert from micrometers to nanometers, we multiply by 1000:

λ ≈ 0.01351 μm * 1000 nm/μm
≈ 13.51 nm

Therefore, the wavelength of the light is approximately 13.51 nanometers.

Now let's move on to part b.

The maximum speed of the emitted electrons can be found using the formula for the kinetic energy of an electron:

K.E. = h * ν - W

where K.E. is the kinetic energy, h is Planck's constant (6.63e-34 Js), ν is the frequency of the light (c/λ), and W is the work function of the metal.

First, we need to find the frequency of the light. We already know the wavelength (λ = 13.51 nm), so we can use the speed of light (c = 3.00e8 m/s) to find the frequency:

ν = c / λ
= 3.00e8 m/s / (13.51e-9 m)

Calculating this, we find:

ν ≈ 2.22e16 Hz

Now we can substitute this value, along with the value for the work function W = 2.28 eV (electron volts), into the kinetic energy formula:

K.E. = h * ν - W
= (6.63e-34 Js) * (2.22e16 Hz) - (2.28 eV * 1.60e-19 J/eV)

Calculating this, we find:

K.E. ≈ 2.93e-19 J

To find the maximum speed of the electrons, we can use the formula:

K.E. = (1/2) * m * v^2

where m is the mass of the electron (9.11e-31 kg) and v is the velocity.

Solving for v:

v^2 = (2 * K.E.) / m
v = sqrt((2 * 2.93e-19 J) / (9.11e-31 kg))

Calculating this, we find:

v ≈ 3.57e6 m/s

Therefore, the maximum speed of the emitted electrons is approximately 3.57 million meters per second.