Physics, 23.12.2019 23:31 lol6998

The following table lists the work functions of a few common metals, measured in electron volts. metal φ(ev) cesium 1.9 potassium 2.2 sodium 2.3 lithium 2.5 calcium 3.2 copper 4.5 silver 4.7 platinum 5.6 using these data, answer the following questions about the photoelectric effect. part a light with a wavelength of 190 nm is incident on a metal surface. the most energetic electrons emitted from the surface are measured to have 4.0 ev of kinetic energy. which of the metals in the table is the surface most likely to be made of? view available hint(s) submit part b of the eight metals listed in the table, how many will eject electrons when a green laser (λg=510nm) is shined on them? view available hint(s) submit part c light with some unknown wavelength is incident on a piece of copper. the most energetic electrons emitted from the copper have 2.7 ev of kinetic energy. if the copper is replaced with a piece of sodium, what will be the maximum possible kinetic energy k of the electrons emitted from this new surface? enter your answer numerically in electron volts to two significant figures. view available hint(s)

Answers: 2

Show answers

Answers

Answer from: babyduckies37

A. Lithium

The equation for the photoelectric effect is:

$E=\phi + K$

where

$E=\frac{hc}{\lambda}$ is the energy of the incident light, with h being the Planck constant, c being the speed of light, and $\lambda$ being the wavelength

$\phi$ is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

$\lambda=190 nm=1.9\cdot 10^{-7}m$ , so the energy of the incident light is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J$

Converting in electronvolts,

$E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV$

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

$\phi = E-K=6.5 eV-4.0 eV=2.5 eV$

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

$\lambda=510 nm=5.1\cdot 10^{-7} m$

So its energy is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J$

Converting in electronvolts,

$E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV$

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: $\phi = 4.5 eV$

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

$E=\phi+K=4.5 eV+2.7 eV=7.2 eV$

Then the copper is replaced with sodium, which has work function of

$\phi = 2.3 eV$

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

$K=E-\phi = 7.2 eV-2.3 eV=4.9 eV$

Answer from: Quest

trig 15.3 cos

Answer from: Quest

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.