Answer:
material work function is 0.956 eV
Explanation:
given data
red wavelength 651 nm
green wavelength 521 nm
photo electrons = 1.50 × maximum kinetic energy
to find out
material work function
solution
we know by Einstein photo electric equation that is
for red light
h ( c / λr ) = Ф + kinetic energy
for green light
h ( c / λg ) = Ф + 1.50 × kinetic energy
now from both equation put kinetic energy from red to green
h ( c / λg ) = Ф + 1.50 × (h ( c / λr ) - Ф)
Ф =( hc / 0.50) × ( 1.50/ λr - 1/ λg)
put all value
Ф =( 6.63 ×
(3 ×
) / 0.50) × ( 1.50/ λr - 1/ λg)
Ф =( 6.63 × (3 × ) / 0.50 ) × ( 1.50/ 651×
- 1/ 521 ×)
Ф = 1.5305 ×
J × ( 1ev / 1.6 × J )
Ф = 0.956 eV
material work function is 0.956 eV
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When light passes from one medium to another, part of it continues on
into the new medium, while the rest of it bounces away from the boundary,
back into the first medium.
The part of the light that continues on into the new medium is <em>transmitted</em>
light. Its forward progress at any point in its journey is <em>transmission</em>.
Its direction usually changes as it crosses the boundary. The bending is <em>
refraction</em>.
The part of the light that bounces away from the boundary and heads back
into the first medium is <em>reflected</em> light. The process of bouncing is <em>reflection</em>.
1) The velocity of the particle is given by the derivative of the position. So, if we derive s(t), we get the velocity of the particle as a function of the time:
2) The acceleration of the particle is given by the derivative of the velocity. So, if we derive v(t), we get the acceleration of the particle as a function of the time:
The problem should only have one part to it, but this one has two.
Before I can do the mass/energy conversion, I have to go and
look up the proton mass for myself ... go out and collect the straw
to make my bricks, as it were. As if the fabulous bounty of 7 points
makes it worth it. They make us do everything around here.
OK. In my Physics book⁽¹⁾, the proton rest mass is
1.67 x 10⁻²⁷ kg.
The formula that relates mass to the equivalent energy is
E = m c² .
The method of applying the formula is known as "plug in what you know",
as follows:
E = (mass) x (speed of light)²
= (1.67 x 10⁻²⁷ kg) x (3 x 10⁸ m/s)²
= (1.67 X 10⁻²⁷ Kg) x (9 x 10¹⁶ m²/s²)
= (1.5 x 10⁻¹⁰) (kg-m²/s²)
= 1.5 x 10⁻¹⁰ joule .
____________________________________
⁽¹⁾ Halliday, David and Resnick, Robert, Physics , John Wiley & Sons,
Inc., 1960, inside front cover, "SELECTED PHYSICAL CONSTANTS".
