Answer:
2577 K
Explanation:
Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.
So, T = ⁴√(P/σεA)
Since P = 60 W, we substitute the vales of the variables into T. So,
T = ⁴√(P/σεA)
= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)
= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)
= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)
= ⁴√(0.00441 × 10¹⁶K⁴)
= 0.2577 × 10⁴ K
= 2577 K
Answer: a) 3.85 days
b) 10.54 days
Explanation:-
Expression for rate law for first order kinetics is given by:

where,
k = rate constant = ?
t = time taken for decomposition = 3 days
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = 
First we have to calculate the rate constant, we use the formula :
Now put all the given values in above equation, we get


a) Half-life of radon-222:


Thus half-life of radon-222 is 3.85 days.
b) Time taken for the sample to decay to 15% of its original amount:
where,
k = rate constant = 
t = time taken for decomposition = ?
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = 


Thus it will take 10.54 days for the sample to decay to 15% of its original amount.
Answer:
Explanation:
Let i be the angle of incidence and r be the angle of refraction .
From the figure
Tan ( 90 - i ) = 2.5 / 8
cot i = 2.5 / 8
Tan i = 8 / 2.5 = 3.2
i = 72.65°
From snell's law
sini / sin r = refractive index
sin 72.65 / sinr = 1.333
sin r = .9545 / 1.333
= .72
r = 46⁰
From the figure
Tan r = d / 4
Tan 46 = d /4
d = 4 x Tan 46
= 4 x 1.0355
=4.14 m .
Answer:
Explanation:
The formula for this, the easy one, is
where No is the initial amount of the element, t is the time in years, and H is the half life. Filling in:
and simplifying a bit:
and
N = 48.0(.0625) so
N = 3 mg left after 12.3 years