In other words a infinitesimal segment dV caries the charge
<span>dQ = ρ dV </span>
<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>
<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>
<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
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
We Know: MA = d₁ / d₂
Here, d₁ = 4
d₂ = 10-4 = 6
Substitute their values into the expression:
MA = 4/6
MA = 2/3
So, Your Answer would be 2/3
Hope this helps!
