Question

4. A photosensitive metal is illuminated by monochromatic light with a wavelength λ = 550 nm. Electrons are emitted due to the photoelectric effect.
(a) If the work function of the metal is φ = 1.8 eV, what is the maximum kinetic energyof an emitted electron? Note: the mass of an electron equals 9.11x10-31 kg.
(b) If the work function of the metal is φ = 1.8 eV, what is the maximum speed of anemitted electron? Note: the mass of an electron equals 9.11x10-31 kg.
(c) What is the threshold (or cutoff) frequency of the metal?
(d) If the photosensitive metal is illuminated with light having a

Answer 1

4. A photosensitive metal is illuminated by monochromatic light with a wavelength λ = 550 nm. Electrons are emitted due to the photoelectric effect.
(a) If the work function of the metal is φ = 1.8 eV, what is the maximum kinetic energyof an emitted electron? Note: the mass of an electron equals 9.11x10-31 kg.
(b) If the work function of the metal is φ = 1.8 eV, what is the maximum speed of anemitted electron? Note: the mass of an electron equals 9.11x10-31 kg.
(c) What is the threshold (or cutoff) frequency of the metal?
(d) If the photosensitive metal is illuminated with light having a frequency equal to thethreshold frequency, what is the theoretical maximum kinetic energy an emittedelectron can have?

Answer 2

@theEric

Answer 3

The experiment is a classical experiment on a phenomenon known as the photoelectric effect. Einstein postulated that light travelled in packets of energy, and this energy is the product of planck's constant h and frequency of the light.
\[E=hf\]
Where E is energy, h is planck's constant and f is frequency.

When the light packet hits the electron, it is absorbed in a all or nothing effect, so it is either absorbed or not. Thus, when the light is absorbed, it acquires the full E which the light had.

With this new energy, the electron is able to escape the metal, so long as it is able to overcome the electrostatic forces of attraction between it and the nucleus, this is known as the work function φ. Thus, the energy that the electron has is:
\[\frac{1}{2}m_{e}v^{2}=E-\phi=hf-\phi\]

Answer 4

what will be A,b,c,d
plz i need help

Answer 5

@radar @Festinger

Answer 6

@hba

Answer 7

i noticed u somehow only seeks the answers.. not the solution. My advice for learning physics would be.. understanding the concept comes 1st, sophisticated equations are merely byproducts of some very basic equations with certain mathematical skills. Sorry no offence, and sry again if im wrong :D

Answer 8

there are reasons why i am only getting answers :(

Answer 9

would you mind helping me @chenping

Answer 10

alright.. using the equation as provided by Festinger:
1/2mv^2=E−ϕ=hf−ϕ (photoelectric equation)
p/s: h is the planck's constant, with a value of 6.626x10^-34
f the freq can be solved as c/入 which is light's speed's constant over lamba(wavelength)

a) as work function is given as 1.8eV (can be converted into joules by multiplying the value of electron charges, 1.602x10^-19 to get 2.8839x10^-19 J
Ke=hf-Work function (W)
= h(c/入）-W
= 6.626x10^-34 . ( 3x10^8/550x10^-9) - 1.8 . 1.602x10^-19
= 7.278x10^-20 J
p/s: mass given for electron is not needed in this qs, but is needed for the following qs.

b) with the same work function given, Kinetic energy equals 1/2mv^2.
Making v (speed) as the title, v=square root (2Ke/m)
v=\[ v=\sqrt{(2x7.278\exp-20/9.11\exp-31)}\]
= 4exp6 ms-1 (x10^anything is actually exponent or shortformed as exp here)

c) Assuming the work function is the same as stated in the 2 qs earlier, equals it to hf
Reversing the position to obtian freq, f=W/h
Thus, f=7.278exp-20/6.625exp-34
= 1.098exp14 Hz

d) 0 because the energy supplied is equavelent to the work function. Ke=1.8eV-1.8eV= 0 J

:D

Answer 11

OMG, This is the most beautiful answer i have ever seen in my life :D

Answer 12

(THANK YOU SO MUCH)^ infinity