-3, because you have to subtract
Answer:
a) n = 9.9 b) E₁₀ = 19.25 eV
Explanation:
Solving the Scrodinger equation for the electronegative box we get
Eₙ = (h² / 8m L²2) n²
where l is the distance L = 1.40 nm = 1.40 10⁻⁹ m and n the quantum number
In this case En = 19 eV let us reduce to the SI system
En = 19 eV (1.6 10⁻¹⁹ J / 1 eV) = 30.4 10⁻¹⁹ J
n = √ (In 8 m L² / h²)
let's calculate
n = √ (8 9.1 10⁻³¹ (1.4 10⁻⁹)² 30.4 10⁻¹⁹ / (6.63 10⁻³⁴)²
n = √ (98) n = 9.9
since n must be an integer, we approximate them to 10
b) We substitute for the calculation of energy
In = (h² / 8mL2² n²
In = (6.63 10⁻³⁴) 2 / (8 9.1 10⁻³¹ (1.4 10⁻⁹)² 10²
E₁₀ = 3.08 10⁻¹⁸ J
we reduce eV
E₁₀ = 3.08 10⁻¹⁸ j (1ev / 1.6 10⁻¹⁹J)
E₁₀ = 1.925 101 eV
E₁₀ = 19.25 eV
the result with significant figures is
E₁₀ = 19.25 eV
Answer:
0.44m/s
Explanation:
drift velocity=I/nAq
diameter 12 gauge
wire=0.081inches=0.081*2.5=0.2025cm radius=0.10125cm area=pi*R^2 =20/8.5*10^22*3.14*0.10125^2*10^-4*1.6*10^-19*
V = 0.44m/s
Answer:
460 g
Explanation:
Heat lost by the warm water = heat gained by the cold water
-mCΔT = mCΔT
-m (4.184 J/g/K) (37°C − 85°C) = (1000 g) (4.184 J/g/K) (37°C − 15°C)
-m (37°C − 85°C) = (1000 g) (37°C − 15°C)
-m (-48°C) = (1000 g) (22°C)
m = 458 g
Rounded to two significant figures, you need a mass of 460 g of water.
